Chapter 08: Chemical Equilibrium

(i) Reversible reactions Definition: The reactions in which reactants are converted into products and products are converted back into reactants, and it proceeds in both directions without change in concentrations of reactants and products are called reversible reactions. Examples 218) + 312(2N3(8) PC-5(g) PC 3(g) + C12(g) Explanation • Reversible reactions are represented by double arrow (F). • Proceed in both forward and reverse directions. • Do not go to completion; instead, they reach a state of balance. • Occur in a closed system where no substance enters or leaves. • The concentrations of reactants and products remain constant. • The rate of the forward reaction equals the rate of the reverse reaction, when they achieve equilibrium. • This state does not mean the reactions have stopped, only that the changes balance out. (il) Irreversible reactions Definition: The reactions in which the reactants are completely consumed and converted into products are called irreversible reactions. Example Combustion of methane CH4+20, →CO, +2H2O+ Heat Formation of water 2H,+O›-2H2O Explanation • Irreversible reactions are represented by single arrow (→>). • Reactants are completely converted to products. • The reaction stops when the limiting reactant is used • They are represented by single arrow. (iii) Macroscopic events Definition: Macroscopic events refer to the phenomena that can be observed with the naked eye without considering the individual particles or molecules involved in the process. Examples of macroscopic properties/events • Color change: Rusting of iron. • Evolution or absorption of heat: Exothermic reactions release heat, endothermic reactions absorb it. • Formation of a precipitate: Solid formed from two solutions, mixing silver nitrate and sodium chloride. • Gas evolution: Bubbles or fizzing, vinegar reacting with baking soda. • Change in volume or pressure: Noticeable in reactions involving gases. • Change in composition: A substance transforms into new products, burning of wood These macroscopic observations are often used to infer what's happening on a molecular (microscopic) level (iv) Microscopic events Definition: Microscopic events refer to the phenomena that cannot be observed with the naked eye. Examples of microscopic events • Molecular collisions: Particles interact and transfer energy. • Breaking and forming of chemical bonds: Fundamental to chemical reactions. • Rearrangement of atoms: Leads to new substances being formed • Electron transfer: In redox reactions, electrons move from one species to another. • Changes in molecular structure or geometry: Affects properties and behavior of substances REVERSIBLE REACTIONS, MICROSCOPIC EVENTS AND DYNAMIC EQUILIBRIUM

Dynamic equilibrium Definition: The state in which rate of forward and rate of reverse reaction are continuous in both directions with similar rates is called dynamic equilibrium. Dynamic nature of chemical equilibrium When a reversible reaction reaches equilibrium, it might appear static from a macroscopic point of view concentrations of reactants and products no longer change visibly. At the microscopic level, the system is still highly dynamic: • Reactant molecules continue to collide and form products: • Product molecules simultaneously revert back to reactants. • The forward and reverse reactions occur at equal rates, so there's no net change in concentration. Reaction of steam and carbon monoxide Forward Reaction: At the start maximum number of collisions between them will occurs as microscopic events. • Equal moles of steam and CO. with high frequency of collisions. • Bonds in CO and H2O are broken, forming H2 and CO2. • The forward reaction rate is maximum initially due to high reactant concentrations. H, 0(g) + CO (g) 2(g) 2(g) Reactant concentrations decrease, so forward rate slows down as the reaction proceeds. • Product concentrations increase, so the reverse reaction begins to speed up. • After the product formation reverse reaction starts so bonds are formed and broken at the same time. • This ongoing molecular activity at equilibrium is define as dynamic, not static. Reverse reaction As the products hydrogen (H2) and carbon dioxide (CO2) accumulate during the forward reaction: Their concentrations increase, leading to: • More frequent collisions between H2 and CO2 molecules. • These are microscopic events where molecules interact at the atomic level. During these collisions • Bonds in H2 and COz are broken. • New bonds are formed to regenerate the reactants: CO and H2O. The forward reaction starts with maximum rate and gradually slows down, whereas at the start, the rate of the reverse reaction is low, gradually increases, and finally becomes constant. • A time comes when both reactions proceed at the same rate. • The reaction at this stage is said to be in chemical equilibrium and the concentration of reactants and products become constant. 12(g) + CO (g) 2(g) + CO2(g) Unless the system is somehow disturbed, no further changes in the concentrations will occur State of chemical equilibrium Definition: The state of a reversible reaction at which composition of the reaction mixture does not change and forward and reverse rates are equal is called the state of chemical equilibrium. Concentration time graphs in reversible reactions Consider the reaction A=-B The concentration of A (reactant) and B (product) is plotted over time. Before equilibrium • [A] (reactant) decreases over time. • [B] (product) increases as it is formed from A. At equilibrium curves • The level off, meaning concentrations of both A and B become constant. • No further net change, even though the forward and reverse reactions continue (dynamic equilibrium). Different scenarios at equilibrium A general reaction in which a reactant A is converted into product B. After attaining the chemical equilibrium, the graph shows three possibilities. (i) Equal concentration: In this type of graph, the reactants and products are in equal concentration at the time of equilibrium. • [A] = [B] at equilibrium. • The system reaches a point where reactant and product are in equal concentrations. (ii) Products are greater than reactants • [A] < [B] at equilibrium. • The reaction favors the products (more B formed) after achieving the equilibrium. (ili) Reactants are greater than products • Concentration of A and B are constant. • [A] > [B] at equilibrium. • The reaction favors the reactants (less B formed) after achieving the equilibrium. (a) [Reactant]=[Product] (b) [Reactant]<[Product] A A Equilibrium Concentration B B Time Time Fig: Plot of concentration vs time • When a reversible reaction reaches equilibrium, the concentrations of reactants and products remain constant over time. • This does not mean the reaction has stopped Microscopic activities • Reactant molecules are still converting into products. • Product molecules are simultaneously converting back into réactants. • These processes are ongoing and balanced. [CO] or [H2O] Equilibrium Concentration Time Fig: Plot of concentration vs time (c) [Reactant]>[Product] A Equilibrium Equilibrium Concentration B Time Equal reaction rates • The rate of the forward reaction equals the rate of the reverse reaction. • There is no net change in concentrations, even though reactions are still happening because these rates are equal. • The system is not static, molecular collisions, bond breaking, and bond formation continues. • Individual molecules are continuously reacting but the rates of forward and reverse reaction are equal. • It's dynamic at the microscopic level, even though macroscopic properties (like color, pressure, or concentration) appear unchanged This concept is essential in understanding how equilibrium systems behave under different conditions.

Relationship between microscopic and macroscopic events • Macroscopic events are the observable changes in a system. These are the result of multiple simultaneous microscopic events at the molecular level. • Importance of understanding microscopic events: Understanding microscopic events helps us to explain and predict the macroscopic changes that occur in an equilibrium system. • Microscopic events in a chemical reaction: When there is a change in equilibrium, it is due to microscopic activities such as collisions between particles or molecules, formation and breaking of chemical bonds. • Factors affecting reaction rates: These collisions affect the rates of the forward and reverse reactions. Reaction rates are influenced by, Activation energy, external factors such as temperature, pressure, and concentration. Observable effects of microscopic changes: The combined effect of these microscopic events causes macroscopic changes. These changes can be observed by the naked eye, such as change in color, formation of precipitates, change in temperature. Quick Check 8.1 (a) Differentiate between macroscopic and microscopic events. Ans. Macroscopic Events •Observable on a scale large (bulk properties) Examples: color change, pressure, volume, temperature Measurable directly using instruments (b) The equilibrium is dynamic in nature, explain in terms, of microscopic events. • At equilibrium, the rate of forward reaction is equal to rate af reverse reaction. • Microscopically, both reactions are still occurring, but their effects cancel out. • So, the concentrations remain constant even though molecules are still reacting. Microscopic Events Occur at the atomic or molecular level Examples: collisions of molecules, bond breaking/forming Not visible; explained using models and theories The system appears static macroscopically, but microscopic changes continue, making equilibrium dynamic in nature. (c) In the reaction H, (g) + CO (g) H 2(g) + CO 2(g) the concentration of the products have higher concentration at equilibrium. (i) Plot a graph between concentrations of reactants and products vs time. (ii) Plot a graph between rate of the reaction with respect to time. Ans. (i) Graph between concentration and time • Initially, reactant concentrations are high and product concentrations are zero. • Over time, reactant concentrations decrease, and product concentrations increase. • Eventually, they reach a point where concentrations become constant. Sketch (ii) Graph between rate and time • Initially, forward rate is high, reverse is zero. • As products form, reverse rate increases, forward rate decreases. • At equilibrium: both rates become equal and constant. ale Sketch a) 4 b) 4 A A Equilibrium Concentration Concentration B B Time Time DYNAMIC EQUILIBRIUM BETWEEN TWO PHYSICAL STATES

Dynamic Equilibrium Definition: The state in which the rate of forward and reverse reaction are continuous in both directions with similar rates is called state of dynamic equilibrium. • Dynamic equilibrium is a state in a reversible process where, the rate of change in one direction is equal to the rate of change in the opposite direction. • Even though continuous changes occur at the microscopic level, there is no net change in the system as a whole. Dynamic equilibrium in phase changes: A reversible phase change also shows dynamic equilibrium between two physical states of a substance. [CO] or [H2O] on all Equilibrium Concentrati Time m c) 4 A : Equilibrium ; Equilibrium Concentration B Time Example: At 0 °C, ice and water coexist in equilibrium, Water freezes into ice and ice melts into water at the same rate. Characteristics • No net change is observed in the amount of each phase. • The macroscopic properties of the system remain constant over time. Relationship between dynamic equilibrium and vapor pressure Studying the behavior of liquids and their vapor pressures is a significant concept in chemical equilibrium. (i) Evaporation and condensation in a closed system • When a liquid is placed in a closed container, some molecules at the surface gain enough kinetic energy to escape into the gaseous phase this is called evaporation. • As more molecules evaporate, the vapor pressure (pressure exerted by gas molecules) increases at the walls of vessel. • Simultaneously, gas molecules may collide with the liquid surface and return to the liquid phase this is called condensation. Contention Gas (i) Establishing dynamic equilibrium • As evaporation and condensation continue, a point is reached where, the rate of evaporation is equal to rate of condensation. This state is called dynamic equilibrium. • At this point, the number of molecules entering and leaving each phase becomes constant. The vapor pressure remains constant, until the temperature is constant and the system is undisturbed (iii) Effect of temperature on vapor pressure • Increasing temperature raises the average kinetic energy of liquid molecules. • Faster the evaporation, higher will be the vapor pressure. • A new dynamic equilibrium is established at the higher temperature with a higher equilibrium vapor pressure. • Dynamic equilibrium in vapor pressure describes the state at which evaporation and condensation occur at equal rates • Vapor pressure remains constant at constant temperature, and increases with temperature. Quick Check 8.2 (a) Dynamic equilibrium exists between water and its vapor at 100°C. Justify. Ans. • At 100°C, water boils, so liquid water and water vapor coexist • Water molecules evaporate into vapor, and vapor molecules condense back into liquid. • At equilibrium, the rate of evaporation = rate of condensation. • Thus, the process, is dynamic (both forward and reverse processes continue), but the amount of liquid and vapor remains constant. (b) Do you think that dynamic equilibrium exists between ice and water at 0°C? If yes, explain. Ams. Yes, at 0°C, ice and water coexist. • Ice melts to water, and water freezes back to ice simultaneously. • The rates of melting and freezing become equal at equilibrium. • So, the system is in dynamic equilibrium with constant amounts of ice and water Conditions for equilibrium To study chemical equilibrium, the following conditions must be fulfilled: (i) Reversibility of reaction Equilibrium occurs only in reversible reactions. • The reaction must proceed in both forward and reverse directions. (ii) Closed system •. The reaction must take place in a closed vessel. • No reactants or products should be allowed to escape from the system. a) CaCO3(s) CaO(s) + CO_(g) Fig: (a) Equilibrium is established when system is closed (b) Equilibrium cannot be established in open system CHARACTERISTICS OF CHEMICAL EQUILIBRIUM

Characteristies of chemical equilibrium Important features of equilibrium are as follows: (i) Concentration: At the stage of chemical equilibrium, the concentrations of reactants and products remain constant. (i) Starting point: The state of equilibrium in a reversible reaction can be approached from either side whether we start with reactants or products. (iii)Role of catalyst: A catalyst does not change the equilibrium position and the equilibrium constant of the reaction. It helps to attain the equilibrium earlier (iv) Dependence of Kc on temperature: The value of equilibrium constant does not depend upon the initial concentrations of reactants, rather it is constants and depends on temperature only. Keep in mind! S.Q. What is the effect of the concentration of pure solid or liquid on equilibrium constant? Ans. If pure solids or pure liquids are involved in an equilibrium system; their concentrations are not included in the equilibrium constant expression. This is because the change in concentration of any pure solid or liquid has no effect on the equilibrium constant. b) CaO(s) + CO_(g) CaCO3(s) Types of equilibrium With respect to the physical states of reactants and products, there are two types of chemical equilibrium. (i) Homogeneous equilibrium Definition: An equilibrium system in which all of the reactants and products are in the same phase. Example: The following are examples of homogeneous equilibria. N2(g) + 3H1 2(g) 2502(g) + 02(g) CH,COOH, + C,H2OH,) - (ii) Heterogeneous equilibrium Definition: An equilibrium in which the reactants and products are in more than one phases is called heterogeneous equilibrium. Example: The following are examples of heterogeneous equilibrium. CaCO 3(s) = CaO(8). + CO2(g) C(s) + H,0 = 00(8) + 1218) Fe, 46) + 4H 2(g) 3Fe (g) + 4H, (8) Quick Check 8.3 (a) Differentiate between homogeneous and heterogeneous equilibria. Ans. Homogeneous Equilibrium All reactants and products are in the same phase (all gases or all liquids or all solids). Example N2(g) + 3H 2(g) = 2NH-3(8). (b) Why do forward reaction rates slow down when a reversible reaction approaches the equilibrium stage? Ans. Initially, forward reaction rate is high; reverse reaction is slow. As products form, reverse reaction rate increases. Near equilibrium, forward and reverse rates become equal. So, the net rate of forward reaction slows down and balances with reverse. (c) In capped soda water bottles, gaseous CO2 is in equilibrium with the aqueous CO2 (HCO3 and Ht). CO2(g) + H2O) → HCO3(a) + H* (i) To which direction will the equilibrium shift if the bottle is opened? (i) Condition for this equilibrium requires the closed cap of the bottle. Why? (iii) When the cap of the bottle is removed, to which direction the equilibrium shifts? Ans. (i) COz escapes when bottle is opened. So direction of reaction is forward and equilibrium shift towards the reactant side (left). (il) The system needs to be closed to prevent escape of gases like CO2. • If the cap is removed, gas escapes and the equilibrium is disturbed. • So, the closed cap maintains a constant volume and pressure, allowing equilibrium to establish. == 2NH3(8) = 2SO 3(g) Heterogeneous Equilibrium Reactants and products are in different phases (solid, liquid, gas). Example CaCO3(9) = CaO(8) + CO2(g) (iii) • Removing the cap allows COz gas to escape. • According to Le Chatelier's Principle, equilibrium shifts to produce more CO2 gas to replace lost gas. So, the equilibrium shifts towards the left (reactants) side. EQUILIBRIUM CONSTANT AND POSITION OF EQUILIBRIUM

Position of Equilibrium • The position of equilibrium refers to the relative amounts of reactants and products present in an equilibrium mixture. • It indicates whether the reactants or products are favored in a reversible reaction. Law of Mass Action Introduction: C. Guldberg and P. Waage, (1864) Norwegian chemists, observed that, for any reversible reaction. Statement: The rate at which reaction proceeds is directly proportional to the product of active masses of the reactants. Equilibrium constant Kc Definition: The ratio of the product of equilibrium concentrations of products (raised to their stoichiometric coefficients) to the product of equilibrium concentrations of reactants (also raised to their coefficients) is constant under given conditions. Consider a general reaction • This expression is known as the law of mass action. • I] denotes molar concentration in mol dm3, also called active mass. • Active mass is the concentration of a substance involved in a reaction. • The rate of a chemical reaction is directly proportional to the product of the active masses of the reactants, each raised to its respective coefficient in the balanced equation. Let us consider a general reaction = C+D AtB - According to the Law of mass action, Rate of forward reaction a [A][B] = kf [A]B] 'kr is the proportionality constant, and is known as forward rate constant. Rate of reverse reaction a [C][D] = kr [C]D] 'kr is the proportionality constant, and is known as reverse rate constant Rate of reverse reaction a [C]D] = kr [C]ID] 'kr is the proportionality constant and is known as reverse rate constant. At the equilibrium stage, the forward and the reverse rates are equal. Hence, kr [A][B] = kr [C][P] K_ ICID] K, A]B The left side of this equation is the ratio of two rate constants, so it gives another constant called the equilibrium constant (Kc). = [CIP] So, [A][B] This equation is known as equilibrium constant expression. General Reaction Consider the following reversible reaction, aA + bB == cC + dD Where a, b, c and d are the coefficients of balanced equation. Then treeim Equilibrium constant expressions of some important reactions 1) N2(g) + 312(g) = = 2NH 3(8) = - c (11) 2N20 5(8) 4 NO 2(g) + 02(g) [N,0,} Quick Check 8.4 (a) Write Kr expressions for the given reactions: Sn?+ Ans. (1) (aq) (29) = Sn+ (ag) + 2Fe3+ (ag) + 2Fe2+ [Sn?][Fe3+1 (ii) 2(g) 4 NH-3(8) + 50 = 4NO(8) + 6H2O(B) INH, 110,1 (ili) PC-5(g) i → PC/3(8) + C218) K=PC, ICL,1 [PC!,] (b) Calculate the value of Ko for the following reaction using the information below: (g) + CO(8) Initial concentrations: [H2 (g)] = 10.00 mol/ dm Equilibrium concentration: [CO (g)] = 9.47 mol/ dm Ans. Step 1: Determine change in concentration • For every mole of CO formed, 1 mole of H2 and 1 mole of CO2 are consumed. x = 9.47 mol/ dm So, Step 2: Calculate equilibrium concentrations of reactants [H, lea = 10.00-9.47 = 0.53 mol/ dm [CO, lea = 10.00-9.47 = 0.53 mol [H, Olea = x = 9.47 mol/ dm Step 3: Write Kc expression: (H2O]ICO] _ (9.47)(9.47) . (0.53)(0.53) Step 4: Calculate Kc 89.7 - = 319.3 K. = 0.2809 Answer Kc = 319.3 Units of equilibrium constant (Kc) (i) Units of equilibrium constant when moles are equal • The units of Kc depend on the number of moles of reactants and products in the balanced chemical equation. • When the total number of moles of reactants equals the total number of moles of products, the units cancel out. • The equilibrium constant Ko becomes dimensionless (has no units). Example: Ester formation reaction CH, COOH (ag) + C,H2OH(ag) = CH, CC, H5(ag) + H, 0(г) K. = [CH, COOH (ag) C, H, OH (aq) ] • Esterification occurs in solution state and follows this pattern: The number of moles of reactants = number of moles of products. • The value of Ko for this reaction has no unit. [CO 2(g)] = 10.00 mol/ dm moldm " x mol dm3 moldm "x mol dm " = no units (ii) Units of kc when moles of reactants and products are unequal • When the number of moles of reactants and products are unequal, the units of the equilibrium constant (Kc) do not cancel out. • The value of kc depends on the units of concentration used (typically mol dm3). Example: Synthesis of ammonia (Haber's process) 2(g) + 312(g) = 2N13(8) [moldm3p [moldm "][mol dm313 = [moldm3] K, = mol-1 dm • Reactant side: 1 + 3 = 4 moles Product side: 2 moles • Change in number of moles: -2 The number of moles are not equal, the units of Ko are mole do RELATIONSHIPS BETWEEN VARIOUS EQUILIBRIUM CONSTANTS

Relationship between equilibrium constants There are four different types of quantities which may be used to calculate the equilibrium constants of reversible reaction. Let the general reaction be aA+ bB = cC + dD (1) Kc When the concentrations of reactants and products are in mole dm , then equilibrium constant is written as: (i) [ATIBI Square brackets [ ] are used for mole dm3 (ії) Кp When the concentrations are expressed in terms of partial pressures (p) for gaseous reactants and products, then (ii) kp= Pc • Ps PA • PB (iii) kn When the concentrations are expressed in terms of number of moles, then n° • n$ (iii) K, = (iv) Kx When the concentrations are expressed in terms of mole fractions, (4) then (iv) Ky = Xa a • XB The relationships between these equilibrium constants are as follows: kp=k. (RT) n Кp =k, (P) An K=k (N) R = General gas constant T= Absolute temperature of the system P = Pressure of the system N = Total number of moles of reactants and products An = Difference of number of moles of products and reactants in the balanced chemical equation • It depends upon the value of 'An' that which of the equilibrium constants is bigger or smaller than the other • If the number of moles of reactants and products in a balanced chemical equation are equal, and all the constants have equal values. An = 0 then Kp = Kc = Kx = Kn Whichever concentration units are used, the equilibrium constants are same. if change in number of moles of reactants and products are equal. Position of equilibrium and reaction conditions The position of equilibrium refers to the relative amounts of products and reactants present in an equilibrium mixture. If a system in equilibrium is disturbed. (i) [Products] > [Reactants] If the concentration of products is increased relative to the reactants, we say that the position of equilibrium has shifted to the left. (il) [Reactants] > [Products] If the concentration of products is decreased relative to the reactants, we say that the position of equilibrium has shifted to the right. Quick Check 8.5 (a) Compare the magnitudes of kc and kp for the following reactions (i) Ammonia synthesis: N2 (g) + 3H, (g) = 2NH, (g) Ans. (i) Change in number of moles of gas: An = moles of products - mole of reactants = 2 - (1 + 3) = - 2 Relationship: < Kc k, = k. (RT)" as (Since An = -2) Kp ka Kp> Kc for dissociation of PCls(g) because the change in number of moles is positive. (b) Nitrogen reacts with hydrogen to form ammonia. = ZNH3(g) 12(0) + 312(g) = The pressure exerted by this mixture of hydrogen, nitrogen and ammonia at constant temperature is 2.0 × 107 Pa. under these conditions, the partial pressure of hydrogen is 0.4 × 107 Pa. calculate the value of Kp for this reaction. Given: Ans. Total Pressure = 2.0 × 10 Pa PN. = 1.5 x 10 Pa PH, = 0.4 x 10 Pa Find PNH, =? PH, = roal - PN, - PH, =2.0x10'-(1.5 + 0.4)x 10' =1.0x10°Pa Write kp expression: (PNH,)? кp =- . Substitute values: (1:0 x 106)2 Kp = (1.5 × 107)(0.4 × 107)3 1.0x1012 кp = 9.6x1026=1.04x 10-15 Answer: Kp= =1.04 × 10-15 LE-CHATELIER'S PRINCIPLE

Le-Chatelier's principle Statement: It can be stated as follows: "If a system in equilibrium is disturbed, it behaves in such a way as to nullify the effect of that disturbance". It describes what happens to a system when something momentarily takes it away from equilibrium. Applications of Le-Chatelier's principle The most common applications of this principle with reference to certain physical and chemical equilibria are discussed below: (i) Effect of change in concentration (ii) Effect of change in pressure (iii) Effect of change in temperature (iv) Effect of catalyst on equilibrium 1. The effect of change of concentrations • When a system is at equilibrium, the concentrations of reactants and products remain constant. • If a reactant or product is added or removed, the equilibrium position is disturbed. • The system then adjusts itself to restore equilibrium this is explained by Le Chatelier's Principle. There are four possibilities for changes in concentration (i) Addition of a reactant (ii) Addition of a product (iii) Removal of a reactant (iv) Removal of a product Hydrolysis of BiCl Consider reversible reaction proceed by combining BiCl, with H2O to give BiOCe (insoluble white solid) and HCl. The kc expression for the reaction is: [BiOC! [HC!] k. = [BiCe: 1[H2O] BiCl, reacts with water to form BiOCl, a white insoluble precipitate, and hydrochloric acid (HCl). If BiOCl is a solid, its concentration is considered constant and may be omitted from the expression depending on the context. • When water is added to BiCl solution, it becomes cloudy due to formation of BiOCe precipitate. • At equilibrium, a certain amount of BiOCl and HCe is formed, some BiCe, remains unreacted. (i) Addition of BiCl, (reactant) • Equilibrium position is disturbed by the addition of BiCl • Equilibrium shifts to the right (forward direction) • More BiOCl and HCl are formed to re-establish the equilibrium. • Kc remains constant, but equilibrium position will change. (ii) Addition of BiOCl or HCe (product) • Equilibrium will be disturbed by the addition of BiOCl. • Equilibrium shifts to the left (reverse direction) • More BiCl, is produced to re-establish the equilibrium. • Kc remains constant, but the position of equilibrium changes. Henry-Louis Le Chatelier (1850-1936) A French chemist proposed the Le-Chatlier principle, a significant achievement in chemistry → BiOCt + 2HCl (ii) Removal of BiCl, (reactant) • Equilibrium will be disturbed by the removal of BiC C3. Equilibrium shifts to the left (reverse direction). • Reaction produces more BiCC, to compensate, and to re-establish the equilibrium. • A new equilibrium is established, kc stays constant. (iv) Removal of BiOCl or HCe (product) • Equilibrium will be disturbed by the removal of BiOCl. • Equilibrium shifts to the right (forward direction). • More BiOCE and HCl are produced to re-establish the equilibrium. • A new equilibrium is reached, with kc unchanged. 2. The effect of change in pressure or volume The effect of pressure or volume change does not apply when: (i) No gaseous components are involved • If no gases are involved in the reaction, pressure or volume changes have no effect on equilibrium. (ii) Equal moles of gaseous reactants and products • In gaseous homogeneous equilibrium, if the number of gaseous moles of reactants and products are equal. n reactants = products • Change in pressure or volume has no effect on the equilibrium position. Effect applies only when gaseous moles of reactants and products are unequal. reactants # products Synthesis of Ammonia (Haber's process) N2 + 3H2(g) === 2NH3(g) • Total moles of reactants are equal to 4. • Total moles of products are equal to 2. • So, number of moles. of reactants are not equal to number of mole of products. (i) Increase in pressure / decrease in volume • The number of moles and volume decrease for the forward reaction. • At equilibrium volume of mixture is less than the total volume at initial stage. • If pressure increase the system shifts to the side with fewer gas moles. • In this case the reaction is shifted to forward direction. (ii) Decrease in pressure / increase in volume • If pressure increase or volume is decreased, the system shifts to the side where the reaction mixture occupy more volume. • Synthesis of ammonia is shifted to reverse direction • More N2 and H2 are formed, ammonia formation decreases. Quick Check 8.6 The change of volume or pressure for the following reactions only changes the equilibrium position but not the equilibrium constant. How the direction of reaction changes for each of the following reactions? Reaction 212(g) +2(g) = Ans. Change: Increasing Pressure (P) • Total moles of gas on left = 3, on right = 2 • Effect: The system will shift to fewer moles to reduce pressure. Equilibrium shifts to the right (towards H2O) (ii) Reaction 200(6) + 02(g) 2002(g) Ans. Change: Increasing Volume (V) • Increasing volume means decreasing pressure • Moles: Left = 3, Right = 2 • Effect: System shifts to more moles to increase pressure. Equilibrium shifts to the left (towards CO and Oz) (iii) Reaction N20 4(8) Ans. Change: Increasing Pressure (P) • Moles: Left = 1, Right = 2 • Effect: System shifts to fewer moles. Equilibrium shifts to the left (towards N204) (iv) Reaction 2502(g) + 02(g) 2503(8) Ans. Change: Increasing Volume (V) • Increasing volume = decreasing pressure • Moles: Left = 3, Right = 2. • Effect: System shifts to more moles. Equilibrium shifts to the left (towards SO and Oz) Reaction Change 52620 1 Pressure 1 Volume 200 + 0, = 200, 1 Pressure N,04 = 2N02 1 Volume 250, + 0, = = 2503 = 2H,° (g) . Shift in Direction → (Right) + (Left) + (Left) + (Left) THE EFFECT OF CHANGE IN TEMPERATURE

1. Effect of Change in temperature • Le-Chatelier's principle helps predict the direction of a chemical reaction when temperature is changed. • Temperature is the only factor that changes the value of equilibrium constant (kc). Changing temperature can disturb equilibrium in two types of reactions. (i) Exothermic Reactions Heat is released: AH° is negative Example 2502(g) + 02(g) = 2503(8) The increase in temperature will shift the reaction from right to left, according to the Le- Chatelier's principle. 2502(g) + 02(g) (a) Increase in temperature • The equilibrium shifts left to right in reverse direction. • [SO3] decreases, and [SO2] [O2] increases. • The value of kc decreases. (b) Decrease in Temperature • [SO3] increases, and [SO2] [O2] decreases. • The value of kc increases. • Production of SO is favoured at lower temperatures. kc values: The preparation of SO is not favored at high temperatures. • At 1000 K kc = 2.8 × 10° • At 298 K kc = 1 × 1026 (ii) Endothermic reactions Heat is absorbed: AH° is positive Example: The equation for the conversion of N204 to NO2 is given below: N, 4(8) = 2NO 2(g) The increase in temperature will shift the reaction from left to right, according to the Le- Chatelier's principle. N2 4(g) = (a) Increase in temperature • The equilibrium shifts right (forward reaction). • [NOz] increases, [N2O4] decreases. • The value of kc increases • Product formation is favored at higher temperatures (b) Decrease in temperature • The equilibrium shifts left (reverse reaction). • [NO2] decreases, [N204] increases. • The value of kc decreases. AH° = -198 kJ = 2SO 3(8) AH° = +57.2 kJ Kc Values • At 273 K Kc = 7.7 × 10-5 • At 373 K kc = 0.4 2. Effect of catalyst on equilibrium Catalyst Definition: A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the reaction. • Does not change the position of equilibrium. • . Does not affect the equilibrium constant (Kc). • Only speeds up the rate at which equilibrium is achieved • Yield remains the same, but equilibrium is achieved faster Working of a Catalyst • Provides a new reaction pathway with lower activation energy (Ea). • This means more reactant molecules have enough energy to overcome the activation barrier. • Reaction proceeds faster in both forward and reverse directions. Energy profile diagram • X-axis: Reaction coordinate • Y-axis: Potential energy • The peak of the curve (activation energy) is lowered in the presence of a catalyst. Comparison Property Uncatalyzed Reaction High Activation Energy (Ea) Slow Reaction Speed Equilibrium Constant Unchanged Equilibrium Position Unaffected INDUSTRIAL APPLICATIONS OF CHEMICAL EQUILIBRIUM

Synthesis of Ammonia by Haber's process The maximum yield of product is to be achieved in chemistry is the basic need of a reaction. Le-Chatelier's principle helps to have maximum yield in industrial synthesis of ammonia. Ammonia synthesis from nitrogen and hydrogen is a reversible process. N2(g) + 32(g) 2N3(g) Industrial importance: Ammonia is an important compound used in fertilizers, explosives, and chemicals. activated state → Energy E, (uncatalyzed reaction) activated st É (catalyzed reaction) Products Reactants → Reaction coordinate Fig: Effect of a catalyst on Reversible Reactions. Catalyzed Reaction Lower Faster Unchanged Unaffected AH° = - 46.11 kJ mol-1 Yield of ammonia: The maximum yield of ammonia can be achieved by following ways. (i) Removal of Ammonia (product removal) • Removal of ammonia from the reaction vessel time to time decreases its concentration, according to Le Chatelier's principle. • The equilibrium shifts forward direction, producing more ammonia. (il) Effect of Pressure • Four moles of reactants combine to give two moles of products, reaction happens with the decreasing volume. • High pressure will shift the equilibrium position to the right to give more and more ammonia. (iii): Effect of Temperature • The reaction is exothermic (releases heat). • Lower temperature favors forward reaction (more ammonia). • The equilibrium shifts forward direction, according to Le-Chatelier's principle. Explanation: From the table 8.1 given below and the figure, then it becomes clear that the effect of pressure and temperature on the yield of ammonia is very prominent. The most complete conversion is 98.3% at 473 K (200°C) and 1000 atmospheric pressure. • The yield is being favoured at 200°C, but the rate of reaction becomes very slow and the process becomes uneconomical. • The temperature is raised to a moderate level i.e. 400°C and a catalyst is used to increase the rate. • If we want to achieve the same rate without a catalyst then we require much higher temperature, which lowers the yield. Optimum industrial conditions The most suitable to get maximum yield of ammonia are: • Pressure between 200-300 atmospheres • Temperature around 673K (400°C) • Pieces of iron crystals present in a fused mixture of MgO, Al2O3 and SiOz as catalyst 100 90 1000 atm 80 - Industrial 600 atm € 70 conditions 60 - 300 atra 50 40 30 - 100 atm 20 - 10 10 atm 0- 200 250 300 350 400 450 500 550 600 650 700 → (°C) ammonia VS. of Fig: Percent yield Temperature (°C). at five different operating pressures. At very high pressure and low temperature (top left, the yield is high, but the rate of formation is low. Industrial conditions (circle) are between 200 and 300 atm at about 400°C. Table 8.1: Effect of Temperature on Kc for Ammonia Synthesis T(K) 200 7.15 x 1015 300 2.69 × 108 400 3.94 × 10' 500 1.72 × 102 600 4.53 x 10° 700 2.96 x 10-1 800 3.96 × 10-2 Preparation of sulphur trioxide To manufacture H2SO4, sulphur trioxide gas is produced from SO2 and O2 in a reversible process. - 2503(g) 2502(g) + 02(g) This is a reversible and exothermic reaction. Table 8.2 Effect of temperature on equilibrium position of SO formation 1. Temperature (°C) Kc Mole% of SO3 Sulphur trioxide is a key intermediate in the manufacture of sulphuric acid (H2SO4). Effect of temperature and pressure • High pressure will shift equilibrium forward, increasing SO yield. • Low temperature will favours formation of SO but slows the reaction. • High temperature will increase rate but reduces SO yield. Effect of Temperature at 1 atm Pressure Observation: As temperature increases, kc decreases, and SO yield drops. Optimum conditions for SO: formation • A mixture of SO2 and O2 (air) is passed over a solid catalyst (V20s). • The exothermic reaction raises the gas temperature to about 600°C. • To increase yield, the gas is recycled at a lower temperature (400-500°C). Conclusion: For industrial production of SO3 • Use a catalyst (V20s) to increase the rate. • Maintain moderate temperature (400-500°C). • Operate at high pressure for better yield. • Recycle the gas to maximize efficiency and conserve energy. Quick Check 8.7 Look at the information given in the table below N 2(g) + 3H2(g) =2 NH 3(8) N20 4(8) = 2 NO 2(g) AH°= -92 kJmol-1 AH° = +57 kJmol = (PNo,)2 PN, (PH,) PN, OA T(K) Kp/atm? T(K) 1.0 × 102 400 200 500 1.6 × 10-1 300 600 3.1 × 10-3 400 700 6.3 × 10-5 500 800 7.9 × 10-6 600 AH° = - 97.9 kJ mol-1 400 700 300 200 500 600 5500 690 55 160 13 98 91 75 46 61 31 2S0 2(g) +02(g) = 2SO 3(g) AH° =-197 kJmol (Pso,)' Kp/atm T(K) Kp/atm-' 1.9 × 10-6 600 3.2 × 103 1.7 × 10-1 700 2.0 × 102 5.1 × 10 800 3.2 × 10 1.5 × 103 6.3 900 1A X 104 1000 2.0 (i) How does the proportion of products in the above systems change as the temperature increases? (ii) Calculate the values of Ke for the reactions at 600K? Ans. (i) According to Le - Chatelier's Principle N2(g) + 3H2(g) = 2N3(6) • Increasing temperature adds heat. • The system will oppose this by favoring the endothermic direction (reverse reaction). • The proportion of ammonia decreases as temperature increases. NO 4(8) = 2 NO 2(g) • Increasing temperature favours forward direction. • The proportion of NO2 increases as temperature increases. 2SO 2(g) + 02(g) = 2503(g) • Increasing temperature favours reverse direction. • The proportion of SO decrease as temperature increases. (ii) Calculation of Kc Values at 600K. → 2NH 3(g) 12(g) + 312(g) = Kp= K. (RT)A 3.1 × 10-3 = Kc (8.314 × 600)? Kc = 30.93 Answer N20 4(g) =2 NO 2(g) Kp = Kc (RT)An 1×10+4 = Kc (8.314 × 600) +1 . Kc = 2.005 Answer 2502(g) + 02(g) 2503(8) Kp= Kc (RT) n 3.2×10+3 = Kc (8.314 × 600)1 Kc = 1.59 × 107 Answer SAMPLE PROBLEMS Sample Problem 8.1 Ethanol reacts with ethanoic acid to form ethyl ethanoate and water CH, COOH (.) + C,H2OH .) CH, COOC, H (0) + H,0 (.) Ethanoic acid Ethanol 500 cm of the reaction mixture at equilibrium contained 0.235 mol of ethanoic acid and 0.0350 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water. Use of this data to calculate a value of Ko for this reaction. Step 1 Write out the balanced chemical equation with the concentrations beneath each substance. CH, COOH (e) + C,H2OH (e) 0.235 mol 0.0350 mol 0.5 dm' 0.5 dm 0.470 mol dm3 0.070 moldm3 Water Ethyl Ethanoate + CH, COOC, H5(6) 0.182 mol 0.182 mol 0.5 dm 0.5 dm 0.364 mol dm 0.364 mol dm3 Step 2 Write the equilibrium constant for this reaction in terms of concentrations. _ (0.364 moldm )(0.364 moldm") = ICH, COOC,H,IIH,01 [CH,COOH]IC, H2OH] (0.470 moldm ) (0.070 mol dm ) Step 3 Substitute the equilibrium concentrations into the expression K. = 4.03 (to 3 significant figures) Step 4 Add the correct units by referring back to the equilibrium expression: The units of mol dm cancel out, so Ko has no units. Therefore, Kc = 4.03. Sample Problem 8.2 Propanone reacts with hydrogen cyanide as follows: HCN(g) CH,COCH3(6) + Hydrogen cyanide Propanone A mixture of 0.0500 mol dm3 propanone and 0.0500 mol dm hydrogen cyanide is left to reach equilibrium at room temperature. At equilibrium the concentration of the product is . Calculate Kc for this reaction. 0.0233 mol dm3 Solution Step 1 Write out the balanced chemical equation with all the data underneath: HCN (g) 0.0500 mol dm " 0.0500 mol dm3 Initial conc. ? Equil. conc. Step 2 Calculate the equilibrium concentrations of the reactants. The chemical equation shows that for every mole of product formed, 1 mole of CH3COCH3 and 1 mole of HCN are consumed. So the equilibrium concentrations are as follows: [CH, COCH,] = 0.0500 - 0.02333 = 0.0267 mol dm3 [HCN] = 0.0500 - 0.0233 = 0.0267 mol dm3 Step 3 Write the equilibrium constant for this reaction in terms of concentrations: _ (CH,)C(OH)(CN)CH,] =. (0.0267 moldm?)(0.0267 moldm) 0.000712 Step 4 Substitute the equilibrium concentrations into the expression K= 32.7 dm' mol-1 Sample Problem 8.3 N2(g) and H2(g) combine to form NH3(g). The value of Kc at 500°C is 6.0 × 102 Calculate the numerical value of Kp for this reaction. Solution: =2NH3(g) N2(g) +31 218) 7 K. = 6.0×10-2 T = 500°C+ 273 = 773K CH, C(OH)(CN) CH3(6) Product - CH, C(OH)(CN) CH3(6) O mol dm3 0.0233 mol dm3 0.0233 (0.0233 moldm?) R= 0.0821 dm atm K 'mol-1 K, =? The formula for conversion of Kc to Kp is, K, = Kc(RT)" An= Number of moles of product - Number of moles of reactants An= 2-4=-2 Substituting these values in the expression R =6.0×102(0.0821 dm atm K"' mol-1× 773K) 2 K=6.0×103(63.5 dm atm mol ")? 0.06 6.0×10-2 - = - 4032.25 (63.5)? K, = 1.49×10- So, the value of Kp is less than Kc. Sample Problem 8.4 In the reaction 02(g) 2502(g) + the equilibrium partial pressures at constant temperature are SO2 = 1.0 x 10° Pa, 02 = 7.0 x 10° Pa, SO3 = 8.0 X 10° Pa. Calculate the value of Kp for this reaction. Solution Step 1: Write the equilibrium expression for the reaction in terms of partial pressures. K Pso, × Pos Step 2: Substitute the equilibrium concentrations into the expression. (8.0x106)3 * (1.0x10) (7.0×10°) = 2503(g) P'SOS - = 9.1×10-° Pa