General Applied Chemistry PDF - CHEM1982 (Winter 2025)

CHEM1982 (Winter 2025) General Applied Chemistry Part I Gases; Thermochemistry; Chemical kinetics and Chemical Equilibrium Instructor: Dr. Mason Lawrence Vapor Pressure and Boiling Point The boiling point of a liquid is the temperature at which the vapor pressure equals the exte...

CHEM1982 (Winter 2025) General Applied Chemistry Part I Gases; Thermochemistry; Chemical kinetics and Chemical Equilibrium Instructor: Dr. Mason Lawrence Vapor Pressure and Boiling Point The boiling point of a liquid is the temperature at which the vapor pressure equals the external pressure. The normal boiling point of a substance is observed at standard atmospheric pressure or 760 torr. As the external pressure on a liquid increases, the boiling point increases. 2 Boiling Point and Intermolecular Forces Since a substance’s vapor pressure is affected by the strength of its intermolecular forces, and since boiling point is related to vapor pressure, a liquid’s boiling point is affected by the strength of the intermolecular forces. In general, the weaker the intermolecular forces are, the lower the boiling point. 3 Iodine Subliming Figure 12.18 4 Phase Diagram for CO2 Figure 12.19 5 Phase Diagram for H2O Figure 12.20 6  Sample Problem 12.5: Problem Using a Phase Diagram to Predict Phase Changes PROBLEM: Use the phase diagram for carbon (shown below) to describe the phase changes that a sample of carbon undergoes during the following: (a) The sample is heated at 10² bar (99 atm) from 1000 K to 5000 K. (b) The sample is then compressed at 5000 K to 10⁶ bar (990,000 atm). 7 The Molecular Basis Of Surface Tension A surface molecule experiences a net attraction downward. This causes a liquid surface to have the smallest area possible. An interior molecule is attracted by others on all sides. Surface tension is the energy required to increase the surface area of a liquid. The stronger the forces between the particles the higher the surface tension. Surface tension decreases with increasing temperature. Figure 12.21 8 Surface Tension, Viscosity, and Forces Between Particles Surface Tension Viscosity Substance Formula (J / m 2 ) at 20°C (N·s / m 2 ) at 20°C Major Force ( s ) Dipole-dipole; Diethyl ether CH 3CH 2OCH 2CH 3 1.7  10−2 0.240  10−3 dispersion Ethanol CH 3CH 2OH 2.3  10−2 1.20  10−3 H bonding −3 H bonding; Butanol CH 3CH 2CH 2CH 2OH 2.5  10 −2 2.95  10 dispersion Water H 2O 7.3  10−2 1.00  10−3 H bonding Mercury Hg 48  10−2 1.55  10−3 Metallic bonding Table 12.3 9 Capillary Action and the Shape Of The Water or Mercury Meniscus In Glass Capillarity is the rising of a liquid through a narrow space against the pull of gravity. A. Water displays a concave meniscus. B. Mercury displays a convex meniscus. Figure 12.22 10 Viscosity Viscosity is the resistance of a fluid to flow, and it results from intermolecular attractions that impede the movement of molecules around and past each other. The stronger the intermolecular forces between particles, the higher the viscosity. Viscosity decreases with increasing temperature. Viscosity is affected by molecular shape; among liquids with the same types of intermolecular forces, those that consist of longer molecules having higher viscosities. 11 Viscosity of Water at Several Temperatures Temperature ( °C ) Viscosity (N  s / m 2 ) * 20 1.00  10−3 40 0.65  10−3 60 0.47  10−3 80 0.35  10−3 *The units of viscosity are newton-seconds per square meter. Table 12.4 12  CHEM1982 (Winter 2025) General Applied Chemistry Part I Gases; Thermochemistry; Chemical kinetics and Chemical Equilibrium Instructor: Dr. Mason Lawrence © McGraw Hill LLC  Phases of Matter Each physical state of matter is a phase, a physically distinct, homogeneous part of a system. The properties of each phase are determined by the balance between the potential and kinetic energy of the particles. The potential energy, in the form of attractive forces, tends to draw particles together. The kinetic energy associated with movement tends to disperse particles. © McGraw Hill LLC 2  A Comparison of Gases, Liquids, and Solids Gas Liquid Solid Kinetic energy dominates the potential Intermolecular attractions between particles are strong Intermolecular forces clearly dominate the kinetic energy of intermolecular forces. enough to pull the particles closer together. energy of the particles. On average, gas particles are far apart Liquid particles touch each other (higher density) and The particles are fixed in position next to each (low density) and move randomly have limited freedom of motion. The particles move other in a regular three-dimensional pattern; solids throughout the container. randomly around each other. have relatively high density and very low freedom of motion; particles in a solid just jiggle in place. The large distance between gas particles Liquids resist an applied force and thus compress only The particles are so close together that a solid means that a gas is highly compressible. slightly. compresses even less than a liquid. Since the gas particles have freedom of The particles still have enough kinetic energy to move Since the particles have very little freedom of motion, a gas fills its container, taking randomly around each other, so a liquid conforms to motion, a solid does not take the shape nor volume both the volume and shape of the the shape of its container, but it has a surface (it does of its container. container. not fill the container). A gas flows easily through another gas. A liquid does flow, but much more slowly than a gas. A solid does not flow significantly. © McGraw Hill LLC 3  Types of Phase Changes and Their Enthalpies Solid to liquid, and vice versa: As the temperature increases, the particles in a solid gain kinetic energy and move out of their fixed positions in the process of melting, or fusion; the opposite change is called freezing. Liquid to gas, and vice versa: As the temperature increases further, the molecules in the liquid phase gain sufficient kinetic energy to separate from each other completely and form a gas in the process of vaporization; the opposite process, changing from a gas to a liquid, is called condensation. Solid to gas, and vice versa: Under certain conditions, as the temperature of a solid increases and the particles gain kinetic energy, they move directly to the gas phase in a process called sublimation; the opposite process is called deposition. © McGraw Hill LLC 4  Heats of Vaporization and Fusion for Several Common Substances © McGraw Hill LLC 5  Phase Changes and Their Enthalpy Changes © McGraw Hill LLC 6  Quantitative Aspects of Phase Changes Within a phase, heat flow is accompanied by a change in temperature, since the average Ek of the particles changes. q = ( amount )  ( heat capacity )  T During a phase change, heat flow occurs at constant temperature, as the average distance between particles changes. q = (amount ) ( H of phase change) © McGraw Hill LLC 7  A Heating-cooling Curve For The Conversion Of Gaseous Water To Ice Figure 12.13 Access the text alternative for slide images. © McGraw Hill LLC 8  Sample Problem Finding the Heat of a Phase Change Depicted by Molecular Scenes PROBLEM: The scenes below represent a phase change of water. find the heat (in kJ) released or absorbed when 24.8 g of H2O undergoes this change. © McGraw Hill LLC 9  Solution © McGraw Hill LLC 10  Liquid-Gas Equilibrium In a closed flask, the system reaches a state of dynamic equilibrium, where molecules are leaving and entering the liquid at the same rate. The vapor pressure is the pressure exerted by the vapor on the liquid. The pressure increases until equilibrium is reached; at equilibrium the vapor pressure is constant. Figure 12.14 © McGraw Hill LLC 11  Factors Affecting Vapor Pressure Vapor pressure is affected by two factors―a change in temperature and a change in the gas itself, that is, in the type and/or strength of intermolecular forces. As temperature increases, the fraction of molecules with enough energy to enter the vapor phase increases, and the vapor pressure increases. higher T  higher P The weaker the intermolecular forces, the more easily particles enter the vapor phase, and the higher the vapor pressure. weaker forces  higher P © McGraw Hill LLC 12  The Effect of Temperature On Vapor Pressure Figure 12.15 Access the text alternative for slide images. © McGraw Hill LLC 13  Vapor Pressure as a Function of Temperature and Intermolecular Forces Vapor pressure increases as temperature increases. Vapor pressure decreases as the strength of the intermolecular forces increases. Figure 12.16 © McGraw Hill LLC 14  Quantifying the Effect of Temperature on Vapor Pressure The Clausius-Clapeyron equation relates vapor pressure to temperature. −H vap  1  ln P =  +C R T  The two-point form is used when the vapor pressures at two different temperatures are known. P2 H vap  1 1  ln =  −  P1 R  T1 T2  R = 8.314 J/mol  K. © McGraw Hill LLC 15  Linear Plots of the Relationship Between Vapor Pressure and Temperature −H vap Slope = R © McGraw Hill LLC 16 Problem PROBLEM: The vapor pressure of ethanol is 119 torr at 34.9°C. If H vap of ethanol is 38.56 kJ/mol, calculate the temperature (in °C) when the vapor pressure is 760 torr. © McGraw Hill LLC 17  Vapor Pressure and Boiling Point The boiling point of a liquid is the temperature at which the vapor pressure equals the external pressure. The normal boiling point of a substance is observed at standard atmospheric pressure or 760 torr. As the external pressure on a liquid increases, the boiling point increases. © McGraw Hill LLC 18  Boiling Point and Intermolecular Forces Since a substance’s vapor pressure is affected by the strength of its intermolecular forces, and since boiling point is related to vapor pressure, a liquid’s boiling point is affected by the strength of the intermolecular forces. In general, the weaker the intermolecular forces are, the lower the boiling point. © McGraw Hill LLC 19  Iodine Subliming Figure 12.18 © McGraw Hill LLC Source: Alexandre Dotta/Science Source 20  Phase Diagram for CO2 Figure 12.19 © McGraw Hill LLC 21  Phase Diagram for H2O Figure 12.20 © McGraw Hill LLC 22 Problem Using a Phase Diagram to Predict Phase Changes PROBLEM: Use the phase diagram for carbon (shown below) to describe the phase changes that a sample of carbon undergoes during the following: (a) The sample is heated at 10² bar (99 atm) from 1000 K to 5000 K. (b) The sample is then compressed at 5000 K to 10⁶ bar (990,000 atm). © McGraw Hill LLC 23  The Molecular Basis Of Surface Tension A surface molecule experiences a net attraction downward. This causes a liquid surface to have the smallest area possible. An interior molecule is attracted by others on all sides. Surface tension is the energy required to increase the surface area of a liquid. The stronger the forces between the particles the higher the surface tension. Surface tension decreases with increasing temperature. Figure 12.21 © McGraw Hill LLC 24  Surface Tension, Viscosity, and Forces Between Particles Surface Tension Viscosity Substance Formula (J / m 2 ) at 20°C (N·s / m 2 ) at 20°C Major Force ( s ) Dipole-dipole; Diethyl ether CH 3CH 2OCH 2CH 3 1.7  10 −2 0.240  10−3 dispersion Ethanol CH 3CH 2OH 2.3  10−2 1.20  10−3 H bonding H bonding; Butanol CH 3CH 2CH 2CH 2OH 2.5  10 −2 2.95  10−3 dispersion Water H 2O 7.3  10−2 1.00  10−3 H bonding Mercury Hg 48  10−2 1.55  10−3 Metallic bonding Table 12.3 © McGraw Hill LLC 25  Capillary Action and the Shape Of The Water or Mercury Meniscus In Glass Capillarity is the rising of a liquid through a narrow space against the pull of gravity. A. Water displays a concave meniscus. B. Mercury displays a convex meniscus. Figure 12.22 © McGraw Hill LLC Source: Stephen Frisch/McGraw Hill 26  Viscosity Viscosity is the resistance of a fluid to flow, and it results from intermolecular attractions that impede the movement of molecules around and past each other. The stronger the intermolecular forces between particles, the higher the viscosity. Viscosity decreases with increasing temperature. Viscosity is affected by molecular shape; among liquids with the same types of intermolecular forces, those that consist of longer molecules having higher viscosities. © McGraw Hill LLC 27  CHEM1982 (Winter 2025) General Applied Chemistry Part I Gases; Thermochemistry; Chemical kinetics and Chemical Equilibrium Instructor: Dr. Mason Lawrence © McGraw Hill LLC Exercise Predicting shifts in Equilibrium For the reaction in which direction will the equilibrium shift when (a) Cl2(g) is removed, (b) the temperature is decreased, (c) the volume is increased, (d) PCl3(g) is added? © McGraw Hill LLC How a System at Equilibrium Responds to Disturbances 3) Effect of Catalysts Catalysts increase the rate of both the forward and reverse reactions. Equilibrium is achieved faster, but the equilibrium composition remains unaltered. Activation energy is lowered, allowing equilibrium to be established at lower temperatures. © McGraw Hill LLC How a System at Equilibrium Responds to Disturbances 3) Effect of Catalysts Exothermic reactions N2 g + 3 H2 g ⇌ 2 NH3 g ∆𝑟 𝐻 𝑜 = −91.8 kJ © McGraw Hill LLC How a System at Equilibrium Responds to Disturbances 3) Effect of Catalysts Exothermic reactions N2 g + 3 H2 g ⇌ 2 NH3 g ∆𝑟 𝐻 𝑜 = −91.8 kJ The Haber process for producing ammonia from the elements is exothermic, cooling down the reactants should result in more product. However, the activation energy for this reaction is high! It is slow of lower temperature This is an instance where a system in equilibrium can be affected by a catalyst, by allowing reactions to reach equilibrium at a lower temperature. © McGraw Hill LLC  Attractive Forces Intramolecular or bonding forces are found within a molecule. The chemical behavior of each phase of matter is the same because the same constituent particle is present in each case. H2O molecules are present whether the substance is in the solid, liquid, or gas phase. Intermolecular or nonbonding forces are found between molecules. The physical behavior of each phase of matter is different because the strength of these forces differs from state to state. © McGraw Hill LLC 6  Intramolecular Forces vs. Intermolecular Forces © McGraw Hill LLC 7  The Nature of Intermolecular Forces Intermolecular forces arise from the attraction between molecules with partial charges, or between ions and molecules. Intermolecular forces are relatively weak compared to bonding forces because they involve smaller charges that are farther apart. © McGraw Hill LLC 8  Covalent and Van Der Waals Radii The van der Waals distance is the distance between two nonbonded atoms in adjacent molecules. The van der Waals radius is one-half the closest distance between the nuclei of two nonbonded atoms. The VDW radius is always larger than the covalent radius. © McGraw Hill LLC 9  Periodic Trends In Covalent And Van Der Waals Radii Like covalent radii (blue quarter-circles and numbers in the figure), van der Waals radii (red quarter-circles and numbers) decrease across a period and increase down a group. © McGraw Hill LLC 10  Comparison of Bonding Forces Basis of Energy Force Model Attraction (kJ/mol) Example Ionic Cation–anion 400–4000 NaCl Nuclei–shared Covalent 150–1100 e⁻ pair Cations– Metallic delocalized 75–1000 Fe electrons © McGraw Hill LLC 11  Comparison of Nonbonding Forces Basis of Energy Force Model Attraction (kJ/mol) Example Ion charge– Ion-dipole 40–600 dipole charge Polar bond to H– dipole charge H bond 10–40 (high EN of N, O, F) Dipole-dipole Dipole charges 5–25 Dispersion Polarizable 0.05–40 (London) e⁻ clouds © McGraw Hill LLC 12  Ion-Dipole and Dipole-Dipole Forces Ion-dipole forces result when an ion and a nearby polar molecule attract each other. The most important example of an ion-dipole force takes place when an ionic compound dissolves in water. Dipole-dipole forces are the attractive forces between the positive pole of one polar molecule and the negative pole of another polar molecule. Dipole-dipole forces only exist for polar molecules. For compounds of similar molar mass, the greater the molecular dipole moment, the greater the dipole-dipole forces. Dipole-dipole forces are not as strong as ion-dipole forces. © McGraw Hill LLC 13  Polar Molecules and Dipole-Dipole Forces © McGraw Hill LLC 14  Dipole Moment and Boiling Point © McGraw Hill LLC 15  The Hydrogen Bond Hydrogen bonding is possible for molecules that have a hydrogen atom covalently bonded to a small, highly electronegative atom with lone electron pairs, specifically N, O, or F. An intermolecular hydrogen bond is the attraction between the H atom of one molecule and a lone pair of the N, O, or F atom of another molecule. Hydrogen bonds are generally stronger than other dipole-dipole forces. Substances that exhibit hydrogen bonding have exceptionally high boiling points. © McGraw Hill LLC 16  Hydrogen Bonding and Boiling Point © McGraw Hill LLC 17  Problem Drawing Hydrogen Bonds Between Molecules of a Substance PROBLEM: Which of these substances exhibits H bonding? Draw examples of the H bonds between two molecules of each substance that does. (a) C2H6 (b) CH3OH (c) © McGraw Hill LLC 18  Dispersion (London) Forces Dispersion forces or London forces arise when an instantaneous dipole in one particle induces a dipole in another, resulting in an attraction between them. Dispersion forces exist between all particles, increasing the energy of attraction in all matter. They are the only force existing between nonpolar particles. Dispersion forces are stronger for more polarizable particles. In general, larger particles experience stronger dispersion forces than smaller ones. As molar mass increases, dispersion forces increase in strength, and so do boiling points. A molecular shape that has more area allows stronger attractions. © McGraw Hill LLC 19  Dispersion Forces Among Nonpolar Particles When atoms are far apart they do not influence one other. When atoms are close together, the instantaneous dipole in one atom induces a dipole in the other. The process occurs throughout the sample. © McGraw Hill LLC 20  Molar Mass and Trends in Boiling Point Dispersion forces are stronger for larger, more polarizable particles. Polarizability correlates closely with molar mass for similar particles. © McGraw Hill LLC 21  Molecular Shape, Intermolecular Contact, and Boiling Point © McGraw Hill LLC 22  Determining the Intermolecular Forces In a Sample © McGraw Hill LLC 23  Problem Identifying the Types of Intermolecular Forces PROBLEM: For each substance, identify the key bonding and/or intermolecular force(s), and predict which substance of the pair has the higher boiling point: (a) MgCl2 or PCl3 (b) CH3NH2 or CH3F (c) CH3OH or CH3CH2OH (d) Hexane (CH3CH2CH2CH2CH2CH3) or 2,2- dimethylbutane. © McGraw Hill LLC 24  Phases of Matter Each physical state of matter is a phase, a physically distinct, homogeneous part of a system. The properties of each phase are determined by the balance between the potential and kinetic energy of the particles. The potential energy, in the form of attractive forces, tends to draw particles together. The kinetic energy associated with movement tends to disperse particles. © McGraw Hill LLC 25  A Comparison of Gases, Liquids, and Solids Gas Liquid Solid Kinetic energy dominates the potential Intermolecular attractions between particles are strong Intermolecular forces clearly dominate the kinetic energy of intermolecular forces. enough to pull the particles closer together. energy of the particles. On average, gas particles are far apart Liquid particles touch each other (higher density) and The particles are fixed in position next to each (low density) and move randomly have limited freedom of motion. The particles move other in a regular three-dimensional pattern; solids throughout the container. randomly around each other. have relatively high density and very low freedom of motion; particles in a solid just jiggle in place. The large distance between gas particles Liquids resist an applied force and thus compress only The particles are so close together that a solid means that a gas is highly compressible. slightly. compresses even less than a liquid. Since the gas particles have freedom of The particles still have enough kinetic energy to move Since the particles have very little freedom of motion, a gas fills its container, taking randomly around each other, so a liquid conforms to motion, a solid does not take the shape nor volume both the volume and shape of the the shape of its container, but it has a surface (it does of its container. container. not fill the container). A gas flows easily through another gas. A liquid does flow, but much more slowly than a gas. A solid does not flow significantly. © McGraw Hill LLC 26  Types of Phase Changes and Their Enthalpies Solid to liquid, and vice versa: As the temperature increases, the particles in a solid gain kinetic energy and move out of their fixed positions in the process of melting, or fusion; the opposite change is called freezing. Liquid to gas, and vice versa: As the temperature increases further, the molecules in the liquid phase gain sufficient kinetic energy to separate from each other completely and form a gas in the process of vaporization; the opposite process, changing from a gas to a liquid, is called condensation. Solid to gas, and vice versa: Under certain conditions, as the temperature of a solid increases and the particles gain kinetic energy, they move directly to the gas phase in a process called sublimation; the opposite process is called deposition. © McGraw Hill LLC 27  Heats of Vaporization and Fusion for Several Common Substances © McGraw Hill LLC 28  Phase Changes and Their Enthalpy Changes © McGraw Hill LLC 29  Quantitative Aspects of Phase Changes Within a phase, heat flow is accompanied by a change in temperature, since the average Ek of the particles changes. q = ( amount )  ( heat capacity )  T During a phase change, heat flow occurs at constant temperature, as the average distance between particles changes. q = (amount ) ( H of phase change) © McGraw Hill LLC 30  A Heating-cooling Curve For The Conversion Of Gaseous Water To Ice Figure 12.13 Access the text alternative for slide images. © McGraw Hill LLC 31  Sample Problem Finding the Heat of a Phase Change Depicted by Molecular Scenes PROBLEM: The scenes below represent a phase change of water. find the heat (in kJ) released or absorbed when 24.8 g of H2O undergoes this change. © McGraw Hill LLC 32  Solution © McGraw Hill LLC 33 CHEM1982 (Winter 2025) General Applied Chemistry Part I Gases; Thermochemistry; Chemical kinetics and Chemical Equilibrium Instructor: Dr. Mason Lawrence Copyright © 2023 Pearson Education, Inc. All Rights Reserved  Reaction Quotient, Q. Predicting the direction of chemical reactions Is a mixture in equilibrium? If not which way does the reaction go? To answer these questions, we calculate the reaction quotient, Q. ⎯⎯ → cC + dD aA + bB ⎯ ⎯ [C]𝑐𝑒𝑞𝑢𝑖𝑙 [D]𝑑𝑒𝑞𝑢𝑖𝑙 [C]c [D]d 𝐾𝑐 = Qc = [A]𝑎𝑒𝑞𝑢𝑖𝑙 [B]𝑏𝑒𝑞𝑢𝑖𝑙 [A]a [B]b The expression for Qc is identical to the expression for Kc, but the values used to calculate it are the current concentrations, often the initial concentrations, not those at equilibrium. Copyright © 2020 Pearson Canada Inc. 14 - 2 Comparing Q and K Both Q and K are numbers that derive from some ratio of the amounts of products over the amounts of reactants. The value of Q relative to K is a measure of the progress of the reaction toward equilibrium. At equilibrium, the reaction quotient is equal to the equilibrium constant. And at equilibrium Qc = Kc Copyright © 2020 Pearson Canada Inc. 14 - 3 Comparing Q and K If Q < K, nature will make the reaction proceed to products. Copyright © 2020 Pearson Canada Inc. 14 - 4 Comparing Q and K If Q < K, nature will make the reaction proceed to products. If Q = K, the reaction is in equilibrium. Copyright © 2020 Pearson Canada Inc. 14 - 5 Comparing Q and K If Q < K, nature will make the reaction proceed to products. If Q = K, the reaction is in equilibrium. If Q > K, nature will make the reaction proceed to reactants. Copyright © 2020 Pearson Canada Inc. 14 - 6 Predicting the Direction of Approach to Equilibrium At 448°C the equilibrium constant Kc for the reaction H2 𝑔 + I2 𝑔 ⇄ 2HI(𝑔) is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0 × 10−2 mol HI 1.0 × 10−2 mol H2 and 3.0 × 10−2 mol I2 in a 2.00 L container. [HI] = 2.0 × 10−2 mol/2.00 L = 1.0 × 10−2 M I2 = 3.0 × 10−2 mol/2.00 L = 1.5 × 10−2 M H2 = 1.0 × 10−2 mol/2.00 L = 5.0 × 10−3 M [HI]2 1.0 × 10−2 2 𝑄𝑐 = = −3 −2 = 1.3 < 50.5 < 𝐾𝑐 H2 I2 5.0 × 10 1.5 × 10 Copyright © 2020 Pearson Canada Inc. 14 - 7 Exercise Predicting the Direction of Approach to Equilibrium Copyright © 2020 Pearson Canada Inc. 14 - 8  Copyright © 2014 by Nelson Education Limited Example: Predicting direction of reaction using Q At 500 K, the reaction I2 (g) ⇌ 2 I (g) has Kp = 5.6 x 10-12. A reaction mixture at 500 K contains PI2 = 0.020 bar and PI = 2.0 x 10-8 bar. Is the mixture at equilibrium? If not, in which direction must the net reaction proceed to reach equilibrium? Copyright © 2020 Pearson Canada Inc. 14 - 9 Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the system will shift its equilibrium position so as to counteract the effect of the disturbance.” After the shift, the system will return to equilibrium. Concentrations and/or pressures will change but the value of the equilibrium constant K remains the same, unless the temperature is change. Copyright © 2020 Pearson Canada Inc. 14 - 10  Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances “If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the system will shift its equilibrium position so as to counteract the effect of the disturbance.” Look at the response of the equilibrium to: changes in the amounts of reactants and/or products changes in the volume or pressure changes in temperature Effect of adding a catalyst The direction of the shift will be such as the compensate for the change, to reduce the disturbance (undo the change). Copyright © 2020 Pearson Canada Inc. 14 - 11 Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances Copyright © 2020 Pearson Canada Inc. 14 - 12 How a System at Equilibrium Responds to Disturbances 1) Adding reactants and/or products The Haber Process The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance. N2 (g) + 3 H2 (g) 2 NH3 (g) Copyright © 2020 Pearson Canada Inc. 14 - 13 How a System at Equilibrium Responds to Disturbances 1) Adding reactants and/or products N2 (g) + 3 H2 (g) 2 NH3 (g) If H2 is added to the system, N2 will be consumed and the two reagents will form more NH3. Adding a reactant shifts the equilibrium towards more products Copyright © 2020 Pearson Canada Inc. 14 - 14 Haber process N2 (g) + 3 H2 (g) 2 NH3 (g) This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid. Copyright © 2020 Pearson Canada Inc. 14 - 15 How a System at Equilibrium Responds to Disturbances 1) Adding reactants and/or products If the system is in equilibrium – adding a reaction component will result in some of it being used up. – removing a reaction component will result in some of it being produced. – magnitude of equilibrium constant remains the same. N2 O4 (g) ⇌ 2 NO2 (g) [NO2 ]2 𝐾𝑐 = [N2 O4 ] Copyright © 2020 Pearson Canada Inc. 14 - 16 Summarizing the Effect of a Concentration Change on Equilibrium If a chemical system is at equilibrium: ▶ Increasing the amount of one or more of the reactants (which makes Q < K ) causes the reaction to shift to the right (in the direction of the products). ▶ Increasing the amount of one or more of the products (which makes Q > K ) causes the reaction to shift to the left (in the direction of the reactants). ▶ Decreasing the amount of one or more of the reactants (which makes Q > K) causes the reaction to shift to the left (in the direction of the reactants). ▶ Decreasing the amount of one or more of the products (which makes Q < K ) causes the reaction to shift to the right (in the direction of the products). Copyright © 2020 Pearson Canada Inc. 14 - 17 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria When gases are involved in an equilibrium, a change in pressure or volume will affect equilibrium: Higher volume or lower pressure favors the side of the equation with more moles of gas (and vice versa). K remains the same. Copyright © 2020 Pearson Canada Inc. 14 - 18 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria N2 g + 3 H2 g ⇌ 2 NH3 g Reaction shifts right Reaction shifts left (toward side with (toward side with fewer moles of gas higher moles of gas particles). particles). Copyright © 2020 Pearson Canada Inc. 14 - 19 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria N2 g + 3 H2 g ⇌ 2 NH3 g At equilibrium at 472 ºC, the partial pressures were PH2 = 7.38 atm PN2 = 2.46 atm and PNH3= 0.166 atm. From these data, the calculated value of Kp was: 2 𝑃NH3 (0.166)2 𝐾𝑝 = 3 = 3 = 2.79 × 10−5 𝑃N2 𝑃H2 (2.46)(7.38) Suppose you cut the volume in half (double the pressures). 2 𝑃NH3 (2 × 0.166)2 −6 𝑄𝑝 = 3 = = 6.97 × 10 𝑃N2 𝑃H2 (2 × 2.46)(2 × 7.38)3 Copyright © 2020 Pearson Canada Inc. 14 - 20 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria N2 O4 (g) ⇌ 2 NO2 (g) Copyright © 2020 Pearson Canada Inc. 14 - 21 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria N2 g + 3H2 g ⇌ 2 NH3 g increase in pressure favours products N2 O4 (g) ⇌ 2 NO2 (g) increase in pressure favours reactants I2 g + H2 g ⇌ 2 HI g increase in pressure has no effect on the equilibrium Copyright © 2020 Pearson Canada Inc. 14 - 22 An example: Kc when one equilibrium concentration is known A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448 C is allowed to reach equilibrium. We calculated that the equilibrium concentrations were 6.5 x 10-5 M, 1.065 x 10-3 and 1.87 x 10-3 M. for H2, I2 and HI respectively, and that Kc at 448 C for this reaction was 50.5 H2 (𝑔) + I2 (𝑔) ⇄ 2 HI (𝑔) 2 1.87×10−3 𝐾𝑐 = = 50.5 6.5×10−5 1.065×10−3 If you double the pressures (concentrations) 2 2 × 1.87×10−3 𝐾𝑐 = = 50.5 2 ×6.5×10−5 2 ×1.065×10−3 Copyright © 2020 Pearson Canada Inc. 14 - 23 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria The partial pressure of the reacting species does not change upon addition of an inert gas. The inert gas does not appear in the equilibrium expression. The position of the equilibrium does not change. (𝑃𝑁𝐻3 )2 N2 g + 3 H2 g ⇌ 2 NH3 g 𝐾𝑝 = (𝑃𝑁2 )(𝑃𝐻2 )3 Copyright © 2020 Pearson Canada Inc. 14 - 24 How a System at Equilibrium Responds to Disturbances 3) Effect of a Temperature Change on Equilibrium Endothermic: Heat acts like a reactant; adding heat drives a reaction toward products. K increases. – Reactants plus heat forms products Exothermic: Heat acts like a product; adding heat drives a reaction toward reactants. K decreases. – Reactants form products plus heat Unlike concentration or pressure changes, changes in temperature change the value of K. Copyright © 2020 Pearson Canada Inc. 14 - 25 How a System at Equilibrium Responds to Disturbances 3) Effect of a Temperature Change on Equilibrium Exothermic reactions N2 g + 3 H2 g ⇌ 2 NH3 g ∆𝑟 𝐻 𝑜 = −91.8 kJ Copyright © 2020 Pearson Canada Inc. 14 - 26 How a System at Equilibrium Responds to Disturbances 3) Effect of a Temperature Change on Equilibrium Endothermic reactions N2 O4 (g) ⇌ 2 NO2 (g) ∆𝑟 𝐻 𝑜 = +57.2 kJ Copyright © 2020 Pearson Canada Inc. 14 - 27 CHEM1982 (Winter 2025) General Applied Chemistry Part I Gases; Thermochemistry; Chemical kinetics and Chemical Equilibrium Instructor: Dr. Mason Lawrence Copyright © 2023 Pearson Education, Inc. All Rights Reserved  Reaction Quotient, Q. Predicting the direction of chemical reactions Is a mixture in equilibrium? If not which way does the reaction go? To answer these questions, we calculate the reaction quotient, Q. ⎯⎯ → cC + dD aA + bB ⎯ ⎯ [C]𝑐𝑒𝑞𝑢𝑖𝑙 [D]𝑑𝑒𝑞𝑢𝑖𝑙 [C]c [D]d 𝐾𝑐 = Qc = [A]𝑎𝑒𝑞𝑢𝑖𝑙 [B]𝑏𝑒𝑞𝑢𝑖𝑙 [A]a [B]b The expression for Qc is identical to the expression for Kc, but the values used to calculate it are the current concentrations, often the initial concentrations, not those at equilibrium. Copyright © 2020 Pearson Canada Inc. 14 - 2 Comparing Q and K Both Q and K are numbers that derive from some ratio of the amounts of products over the amounts of reactants. The value of Q relative to K is a measure of the progress of the reaction toward equilibrium. At equilibrium, the reaction quotient is equal to the equilibrium constant. And at equilibrium Qc = Kc Copyright © 2020 Pearson Canada Inc. 14 - 3 Comparing Q and K If Q < K, nature will make the reaction proceed to products. Copyright © 2020 Pearson Canada Inc. 14 - 4 Comparing Q and K If Q < K, nature will make the reaction proceed to products. If Q = K, the reaction is in equilibrium. Copyright © 2020 Pearson Canada Inc. 14 - 5 Comparing Q and K If Q < K, nature will make the reaction proceed to products. If Q = K, the reaction is in equilibrium. If Q > K, nature will make the reaction proceed to reactants. Copyright © 2020 Pearson Canada Inc. 14 - 6 Predicting the Direction of Approach to Equilibrium At 448°C the equilibrium constant Kc for the reaction H2 𝑔 + I2 𝑔 ⇄ 2HI(𝑔) is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0 × 10−2 mol HI 1.0 × 10−2 mol H2 and 3.0 × 10−2 mol I2 in a 2.00 L container. [HI] = 2.0 × 10−2 mol/2.00 L = 1.0 × 10−2 M I2 = 3.0 × 10−2 mol/2.00 L = 1.5 × 10−2 M H2 = 1.0 × 10−2 mol/2.00 L = 5.0 × 10−3 M [HI]2 1.0 × 10−2 2 𝑄𝑐 = = −3 −2 = 1.3 < 50.5 < 𝐾𝑐 H2 I2 5.0 × 10 1.5 × 10 Copyright © 2020 Pearson Canada Inc. 14 - 7 Exercise Predicting the Direction of Approach to Equilibrium Copyright © 2020 Pearson Canada Inc. 14 - 8  Copyright © 2014 by Nelson Education Limited Example: Predicting direction of reaction using Q At 500 K, the reaction I2 (g) ⇌ 2 I (g) has Kp = 5.6 x 10-12. A reaction mixture at 500 K contains PI2 = 0.020 bar and PI = 2.0 x 10-8 bar. Is the mixture at equilibrium? If not, in which direction must the net reaction proceed to reach equilibrium? Copyright © 2020 Pearson Canada Inc. 14 - 9 Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the system will shift its equilibrium position so as to counteract the effect of the disturbance.” After the shift, the system will return to equilibrium. Concentrations and/or pressures will change but the value of the equilibrium constant K remains the same, unless the temperature is change. Copyright © 2020 Pearson Canada Inc. 14 - 10  Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances “If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the system will shift its equilibrium position so as to counteract the effect of the disturbance.” Look at the response of the equilibrium to: changes in the amounts of reactants and/or products changes in the volume or pressure changes in temperature Effect of adding a catalyst The direction of the shift will be such as the compensate for the change, to reduce the disturbance (undo the change). Copyright © 2020 Pearson Canada Inc. 14 - 11 Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances Copyright © 2020 Pearson Canada Inc. 14 - 12 How a System at Equilibrium Responds to Disturbances 1) Adding reactants and/or products The Haber Process The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance. N2 (g) + 3 H2 (g) 2 NH3 (g) Copyright © 2020 Pearson Canada Inc. 14 - 13 How a System at Equilibrium Responds to Disturbances 1) Adding reactants and/or products N2 (g) + 3 H2 (g) 2 NH3 (g) If H2 is added to the system, N2 will be consumed and the two reagents will form more NH3. Adding a reactant shifts the equilibrium towards more products Copyright © 2020 Pearson Canada Inc. 14 - 14 Haber process N2 (g) + 3 H2 (g) 2 NH3 (g) This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid. Copyright © 2020 Pearson Canada Inc. 14 - 15 How a System at Equilibrium Responds to Disturbances 1) Adding reactants and/or products If the system is in equilibrium – adding a reaction component will result in some of it being used up. – removing a reaction component will result in some of it being produced. – magnitude of equilibrium constant remains the same. N2 O4 (g) ⇌ 2 NO2 (g) [NO2 ]2 𝐾𝑐 = [N2 O4 ] Copyright © 2020 Pearson Canada Inc. 14 - 16 Summarizing the Effect of a Concentration Change on Equilibrium If a chemical system is at equilibrium: ▶ Increasing the amount of one or more of the reactants (which makes Q < K ) causes the reaction to shift to the right (in the direction of the products). ▶ Increasing the amount of one or more of the products (which makes Q > K ) causes the reaction to shift to the left (in the direction of the reactants). ▶ Decreasing the amount of one or more of the reactants (which makes Q > K) causes the reaction to shift to the left (in the direction of the reactants). ▶ Decreasing the amount of one or more of the products (which makes Q < K ) causes the reaction to shift to the right (in the direction of the products). Copyright © 2020 Pearson Canada Inc. 14 - 17 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria When gases are involved in an equilibrium, a change in pressure or volume will affect equilibrium: Higher volume or lower pressure favors the side of the equation with more moles of gas (and vice versa). K remains the same. Copyright © 2020 Pearson Canada Inc. 14 - 18 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria N2 g + 3 H2 g ⇌ 2 NH3 g Reaction shifts right Reaction shifts left (toward side with (toward side with fewer moles of gas higher moles of gas particles). particles). Copyright © 2020 Pearson Canada Inc. 14 - 19 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria N2 g + 3 H2 g ⇌ 2 NH3 g At equilibrium at 472 ºC, the partial pressures were PH2 = 7.38 atm PN2 = 2.46 atm and PNH3= 0.166 atm. From these data, the calculated value of Kp was: 2 𝑃NH3 (0.166)2 𝐾𝑝 = 3 = 3 = 2.79 × 10−5 𝑃N2 𝑃H2 (2.46)(7.38) Suppose you cut the volume in half (double the pressures). 2 𝑃NH3 (2 × 0.166)2 −6 𝑄𝑝 = 3 = = 6.97 × 10 𝑃N2 𝑃H2 (2 × 2.46)(2 × 7.38)3 Copyright © 2020 Pearson Canada Inc. 14 - 20 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria N2 O4 (g) ⇌ 2 NO2 (g) Copyright © 2020 Pearson Canada Inc. 14 - 21 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria N2 g + 3H2 g ⇌ 2 NH3 g increase in pressure favours products N2 O4 (g) ⇌ 2 NO2 (g) increase in pressure favours reactants I2 g + H2 g ⇌ 2 HI g increase in pressure has no effect on the equilibrium Copyright © 2020 Pearson Canada Inc. 14 - 22 An example: Kc when one equilibrium concentration is known A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448 C is allowed to reach equilibrium. We calculated that the equilibrium concentrations were 6.5 x 10-5 M, 1.065 x 10-3 and 1.87 x 10-3 M. for H2, I2 and HI respectively, and that Kc at 448 C for this reaction was 50.5 H2 (𝑔) + I2 (𝑔) ⇄ 2 HI (𝑔) 2 1.87×10−3 𝐾𝑐 = = 50.5 6.5×10−5 1.065×10−3 If you double the pressures (concentrations) 2 2 × 1.87×10−3 𝐾𝑐 = = 50.5 2 ×6.5×10−5 2 ×1.065×10−3 Copyright © 2020 Pearson Canada Inc. 14 - 23 How a System at Equilibrium Responds to Disturbances 2) Effect of Volume and Pressure Changes on Equilibria The partial pressure of the reacting species does not change upon addition of an inert gas. The inert gas does not appear in the equilibrium expression. The position of the equilibrium does not change. (𝑃𝑁𝐻3 )2 N2 g + 3 H2 g ⇌ 2 NH3 g 𝐾𝑝 = (𝑃𝑁2 )(𝑃𝐻2 )3 Copyright © 2020 Pearson Canada Inc. 14 - 24 How a System at Equilibrium Responds to Disturbances 3) Effect of a Temperature Change on Equilibrium Endothermic: Heat acts like a reactant; adding heat drives a reaction toward products. K increases. – Reactants plus heat forms products Exothermic: Heat acts like a product; adding heat drives a reaction toward reactants. K decreases. – Reactants form products plus heat Unlike concentration or pressure changes, changes in temperature change the value of K. Copyright © 2020 Pearson Canada Inc. 14 - 25 How a System at Equilibrium Responds to Disturbances 3) Effect of a Temperature Change on Equilibrium Exothermic reactions N2 g + 3 H2 g ⇌ 2 NH3 g ∆𝑟 𝐻 𝑜 = −91.8 kJ Copyright © 2020 Pearson Canada Inc. 14 - 26 How a System at Equilibrium Responds to Disturbances 3) Effect of a Temperature Change on Equilibrium Endothermic reactions N2 O4 (g) ⇌ 2 NO2 (g) ∆𝑟 𝐻 𝑜 = +57.2 kJ Copyright © 2020 Pearson Canada Inc. 14 - 27 CHEM1982 (Winter 2024) General Applied Chemistry Part I Gases; Thermochemistry; Chemical kinetics and Chemical Equilibrium Instructor: Dr. Mason Lawrence K and Chemical Equation The equilibrium constant of a reaction in the reverse direction is the reciprocal of the equilibrium constant of the forward reaction. [C]3 ⎯⎯ → K forward = A + 2 B ⎯ ⎯ 3C [A][B]2 [A][B]2 1 ⎯⎯ → A+2 B 3 C ⎯ ⎯ K reverse = 3 = [C] K forward For example: NO2 2 N2 O4 𝑔 ⇌ 2 NO2 𝑔 𝐾𝑐 = = 0.212 at 100 °C N2 O4 N2 O4 2 NO2 𝑔 ⇌ N2 O4 𝑔 𝐾𝑐 = = 4.72 at 100 °C NO2 2 1/0.212 = 4.72 Copyright © 2023 Pearson Education, Inc. All Rights Reserved K and Chemical Equation The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. [C]3 ⎯⎯ → K= A + 2 B ⎯ ⎯ 3 C [A][B]2 [C]3n ⎯⎯ → 3n C n A + 2n B ⎯ ⎯ K= = K n [A]n [B]2n For example: NO2 2 1 N2 O4 𝑔 ⇌ 2 NO2 𝑔 𝐾𝑐 = = 0.212 N2 O4 NO2 4 2 2 N2 O4 𝑔 ⇌ 4 NO2 𝑔 𝐾𝑐 = = 0.212 = 0.0449 N2 O4 2 Copyright © 2023 Pearson Education, Inc. All Rights Reserved K and Chemical Equation The equilibrium constant for a reaction that is the sum of one or more reaction steps is the product of the equilibrium constants for the individual reactions + = = 𝐾1 × 𝐾2 K and Chemical Equation The equilibrium constant for a reaction that is the sum of one or more reaction steps is the product of the equilibrium constants for the individual reactions Example: 2 𝑁𝑂𝐵𝑟 ⇄ 2𝑁𝑂 + 𝐵𝑟2 𝐾1 = 0.014 𝐵𝑟2 + 𝐶𝑙2 ⇄ 2 𝐵𝑟𝐶𝑙 𝐾2 = 7.2 2 𝑁𝑂𝐵𝑟 + 𝐶𝑙2 ⇄ 2 𝑁𝑂 + 2 𝐵𝑟𝐶𝑙 𝐾3 = 𝐾1 × 𝐾2 𝐾3 = 𝐾1 × 𝐾2 = 0.014 × 7.2 = 0.10 Copyright © 2023 Pearson Education, Inc. All Rights Reserved Exercise Chemical Kinetics © 2015 Pearson Education, Inc. Heterogeneous Equilibria Homogeneous equilibria occur when all reactants and products are in the same phase. Heterogeneous equilibria occur when a component in the equilibrium is in a different phase. Whenever a pure solid (s) or pure liquid (l) is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium-constant expression. Activity of solids and pure liquids is one (aliquid = asolid = 1). PbCl2 (s) Pb2+ (aq) + 2 Cl-(aq) Kc = [Pb2+] [Cl-]2 Copyright © 2023 Pearson Education, Inc. All Rights Reserved Heterogeneous Equilibria The equation for the reaction is CaCO3 𝑠 ⇌ CaO 𝑠 + CO2 𝑔 𝐾𝑐 = CO2 and 𝐾𝑝 = 𝑃CO2 As long as some CaCO3 and CaO remain in the system, the amount of CO2 above the solid (pressure of CO2) will remain the same. Copyright © 2023 Pearson Education, Inc. All Rights Reserved Example: Analyzing a Heterogeneous Equilibrium Each of these mixtures was placed in a closed container and allowed to stand: (a) CaCO3 (s ) (b) CaO(s ) and CO2 (g ) at a pressure greater than the value of K p (c) CaCO3 (s ) and CO2 (g ) at a pressure greater than the value of K p (d) CaCO3 (s ) and CaO(s ) Determine whether or not each mixture can attain the equilibrium CaCO3 (𝑠) ⇌ CaO(𝑠) + CO2 (𝑔) Copyright © 2023 Pearson Education, Inc. All Rights Reserved Practice Exercise Writing Equilibrium-Constant Expressions for Heterogeneous Reactions Write the following equilibrium-constant expressions: 𝐾𝑐 for Cr(𝑠) + 3Ag + (𝑎𝑞) ⇄ Cr 3+ (𝑎𝑞) + 3 Ag( 𝑠), 𝐾𝑝 for 3Fe(𝑠) + 4H2 O(𝑔) ⇄ Fe3 O4 (𝑠) + 4H2 (𝑔). Copyright © 2023 Pearson Education, Inc. All Rights Reserved Calculating Equilibrium Constants a) Calculating K when equilibrium concentrations are known (have been measured). After a mixture of hydrogen and nitrogen gases in a reaction vessel is allowed to attain equilibrium at 472 ºC, contain 7.38 atm of H2, 2.46 atm N2 and 0.166 atm NH3. From these data, calculate the equilibrium constant Kp the reaction N2 (𝑔) + 3H2 (𝑔) ⇌ 2NH3 (𝑔) ( P ) 2 NH3 (0.166)2 −5 Kp = = = 2.79  10 (P ) (P ) 3 (2.46)(7.38)3 N2 H2 Copyright © 2023 Pearson Education, Inc. All Rights Reserved Calculating Equilibrium Constants b) Calculating K from initial concentrations when only one equilibrium concentrations is known. 1) Tabulate all known initial and equilibrium concentrations. 2) Where the initial and equilibrium concentrations are known, calculate the change. 3) Use the balanced equation to find change for all other reactants and products. 4) Use initial concentrations and changes to find equilibrium concentration of all species. 5) Calculate the equilibrium constant using the equilibrium concentrations. Copyright © 2023 Pearson Education, Inc. All Rights Reserved b) Calculating K from initial concentrations when only one equilibrium concentrations is known. ⎯⎯ → 2 B (g) [B]2 A( g ) ⎯ ⎯ 𝐾c = [A] (I.C.E. tables) [A] [B] Initial 1.00 0 Change Equilibrium 0.75 Copyright © 2023 Pearson Education, Inc. All Rights Reserved An example: Kc when one equilibrium concentration is known A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448 C for this reaction? H2 (𝑔) + I2 (𝑔) ⇄ 2 HI (𝑔) Hl, M Blank H 2  , M left bracket H sub 2 right bracket, M l 2  , M left bracket I sub 2 right bracket, M left bracket H I right bracket, M Initially Change At equilibrium Copyright © 2023 Pearson Education, Inc. All Rights Reserved An example: Kc when one equilibrium concentration is known A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448 C for this reaction? H2 (𝑔) + I2 (𝑔) ⇄ 2 HI (𝑔) Copyright © 2023 Pearson Education, Inc. All Rights Reserved Example: K from Initial and Equilibrium Concentrations Sulfur trioxide decomposes at high temperature in a sealed container according to: Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K. Copyright © 2020 Pearson Canada Inc. 14 - 16  What Do We Know? PSO3 atm PSO2 atm PO2 atm Initial Change Equilibrium Copyright © 2023 Pearson Education, Inc. All Rights Reserved Some Applications of Equilibrium Constants Calculating Equilibrium Concentrations If you know the equilibrium constant, you can find equilibrium concentrations from initial concentrations and changes (based on stoichiometry). You will set up a table (I.C.E. table) similar to the ones used to find the equilibrium concentration, but the “change in concentration” row will be a factor of “x” based on the stoichiometry of the reaction. Then solve for “x” using the known value of K and the equilibrium expression for the reaction. Copyright © 2023 Pearson Education, Inc. All Rights Reserved An Equilibrium Concentration Calculation Example A 𝑎𝑞 ⇄ 2 B 𝑎𝑞 𝐾𝑐 = 0.33 [B]2 (2x)2 4x 2 + 𝐾c x − 𝐾c = 0 𝐾c = = [A] (1.0 − x) 4x 2 + 0.33x − 0.33 = 0 Copyright © 2023 Pearson Education, Inc. All Rights Reserved An Equilibrium Concentration Calculation Example A 𝑎𝑞 ⇄ 2 B 𝑎𝑞 𝐾𝑐 = 0.33 4x 2 + 0.33x − 0.33 = 0 −𝑏 ± 𝑏2 − 4𝑎𝑐 𝑥= 2𝑎 −0.33 ± (0.33)2 −4(4)(−0.33) 𝑥= 2(4) 𝑥 = 0.249 𝑀 𝑜𝑟 − 0.331 𝑀 At equilibrium [A] = 1.0 – 0.249 = 0.751 M [B] = 2x = 0.498 M (0.498)2 Check: 𝐾𝑐 = = 0.33 (0.761) Copyright © 2020 Pearson Canada Inc. 14 - 20 An Equilibrium Concentration Calculation Example: A 1.000 L flask is filled with 1.000 mol of H2 ( g ) and 2.000 mol of l 2 ( g ) at 448 C. Given a K c of 50.5 at 448 degree Celsius, what are the equilibrium concentrations of H2 , l 2 , and HI? H2 (𝑔) + I2 (𝑔) ⇌ 2HI(𝑔) Initial concentration (M) 1.000 2.000 0 Change in concentration (M) −x −x negative x negative x +2 X 1.000 − x 2.000 − x 1.000 minus x Equilibrium concentration (M) 2.000 minus x 2X HI 2 2𝑥 2 𝐾𝑐 = = = 50.5 H2 I2 1.000 − 𝑥 2.000 − 𝑥 Copyright © 2023 Pearson Education, Inc. All Rights Reserved An Equilibrium Concentration Calculation 46.5𝑥 2 − 151.5𝑥 + 101 = 0 −𝑏 ± 𝑏2 − 4𝑎𝑐 𝑥= 2𝑎 −151.5 ± (−151.5)2 −4(46.5)(101) 𝑥= 2(46.5) 𝑥 = 2.323 𝑀 or 0.935 𝑀 H2 eq = 1.000 − 0.935 = 0.065 𝑀 l2 eq = 2.000 − 0.935 = 1.065 𝑀 Hl eq = 2(0.935) = 1.87 𝑀 1.87 2 Check: 𝐾𝑐 = = 50.5 0.065 1.065 Copyright © 2023 Pearson Education, Inc. All Rights Reserved An Equilibrium Concentration Calculation Special cases: When x is small compared to the initial concentrations (when K is small). A 𝑎𝑞 ⇄ 2 B 𝑎𝑞 𝐾𝑐 = 3.3 × 10−5 [B]2 (2x)2 𝐾c = = 4x 2 + 𝐾c x − 𝐾c = 0 [A] (1.0 − x) Copyright © 2023 Pearson Education, Inc. All Rights Reserved  An Equilibrium Concentration Calculation Special cases: When x is small compared to the initial concentrations (when K is small). A 𝑎𝑞 ⇄ 2 B 𝑎𝑞 𝐾𝑐 = 3.3 × 10−5 Assume 𝑥 ≪ 1.0 𝑠𝑢𝑐ℎ 1.0 − 𝑥 ≈ 1.0 [B]2 (2x)2 𝐾c = = 3.3 × 10−5 = [A] (1.0 − x) (2x)2 3.3 × 10−5 ≈ = 4𝑥2 1.0 4x 2 + 3.3 x 10−5 x − 3.3 × 10−5 = 0 𝑥 = 0.002872 𝑀 −3.3 × 10−5 ± (3.3 × 10−5 )2 −4(4)(−3.3 × 10−5 ) 𝑥= 2(4) 𝑥 = 0.002868 𝑀 𝑜𝑟 − 0.00288 𝑀 Copyright © 2023 Pearson Education, Inc. All Rights Reserved An Equilibrium Concentration Calculation Verify the assumption: is x K, nature will make the reaction proceed to reactants. Copyright © 2023 Pearson Education, Inc. All Rights Reserved Predicting the Direction of Approach to Equilibrium At 448°C the equilibrium constant Kc for the reaction H2 𝑔 + I2 𝑔 ⇄ 2HI(𝑔) is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0 × 10−2 mol HI 1.0 × 10−2 mol H2 and 3.0 × 10−2 mol I2 in a 2.00 L container. [HI] = 2.0 × 10−2 mol/2.00 L = 1.0 × 10−2 M I2 = 3.0 × 10−2 mol/2.00 L = 1.5 × 10−2 M H2 = 1.0 × 10−2 mol/2.00 L = 5.0 × 10−3 M [HI]2 1.0 × 10−2 2 𝑄𝑐 = = −3 −2 = 1.3 < 50.5 < 𝐾𝑐 H2 I2 5.0 × 10 1.5 × 10 Copyright © 2023 Pearson Education, Inc. All Rights Reserved CHEM1982 (Winter 2025) General Applied Chemistry Part I Gases; Thermochemistry; Chemical kinetics and Chemical Equilibrium Instructor: Dr. Mason Lawrence The Concept of Chemical Equilibrium Equilibrium is a state of balance. When a reaction is at equilibrium: Concentrations of products and reactants no longer are changing with time. But the reaction has not stopped, it is a dynamic equilibrium Rate𝑓 = Rate𝑟 Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. Copyright © 2023 Pearson Education, Inc. All Rights Reserved  Chemical Equilibrium are Dynamic Equilibrium http://utenti.quipo.it/base5/misure/bilanciainf.htm Static Equilibrium Dynamic Equilibrium Copyright © 2023 Pearson Education, Inc. All Rights Reserved The Concept of Chemical Equilibrium Equilibrium is a state of balance. Rate𝑓 = Rate𝑟 N2O4(g) 2 NO2(g) Copyright © 2023 Pearson Education, Inc. All Rights Reserved The Concept of Chemical Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate (vertical dashed line). Once equilibrium is achieved, the amount of each reactant and product remains constant. Copyright © 2023 Pearson Education, Inc. All Rights Reserved The Equilibrium Constant Dynamic equilibrium for a chemical reaction is the condition in which the rate of the forward reaction equals the rate of the reverse reaction.. The equation is written with a double arrow: N2 O4 𝑔 2 NO2 𝑔 The rate for the forward reaction N2 O4 𝑔 → 2 NO2 𝑔 Rate = 𝑘𝑓 N2 O4 The rate for the reverse reaction 2 NO2 𝑔 → N2 O4 𝑔 2 Rate = 𝑘𝑟 NO2 Copyright © 2023 Pearson Education, Inc. All Rights Reserved Equilibrium Constant Expression At equilibrium 𝑅𝑎𝑡𝑒𝑓 = 𝑅𝑎𝑡𝑒𝑟 𝑘𝑓 N2 O4 = 𝑘𝑟 NO2 2 Rewriting, it becomes the expression for the equilibrium constant (K): 𝑘𝑓 NO2 2 = = 𝐾𝑒𝑞 𝑘𝑟 N2 O4 Specifically, this equilibrium expression is referred to as 𝑲𝒄 where the subscript refers to quantities of reactants and products given as molar concentrations. Copyright © 2023 Pearson Education, Inc. All Rights Reserved  Evaluating 𝐾𝑐 N2 O4 𝑔 2 NO2 𝑔 Table Initial and Equilibrium Concentrations of N2 O4 𝑔 and NO2 𝑔 at 100 ºC Initial Initial Equilibrium Equilibrium N2O 4  ( M ) NO 2  ( M ) N2O 4  ( M ) NO 2  ( M ) Kc Experiment left bracket N sub 2 O sub 4, right bracket, in moles left bracket N O sub 2, right bracket, in moles left bracket N sub 2 O sub 4, right bracket, in moles left bracket N O sub 2, right bracket, in moles K sub c 1 0.0 0.0200 0.00140 0.0172 0.211 2 0.0 0.0300 0.00280 0.0243 0.211 3 0.0 0.0400 0.00452 0.0310 0.213 4 0.0200 0.0 0.00452 0.0310 0.213 NO2 2 𝐾𝑐 = N2 O4 Copyright © 2023 Pearson Education, Inc. All Rights Reserved Equilibrium Constant Expression For any reaction, equilibrium will be reach when ratio of the concentrations the products over concentration of the reactants reaches values such that rate of the forward reaction = rate of the reverse reaction. k1 ⎯⎯→ c C + d D a A + b B ⎯⎯ k −1 [C]𝑐 [D]𝑑 Law of mass action 𝐾𝑐 = [A]𝑎 [B]𝑏 The equilibrium constant is the ratio of the concentration of products over reactants raise to the power of the reaction stoichiometric coefficients. Copyright © 2020 Pearson Canada Inc. 14 - 9 Equilibrium Constant Expression For any reaction, equilibrium will be reach when ratio of the concentrations the products over concentration of the reactants reaches values such that rate of the forward reaction = rate of the reverse reaction. k1 ⎯⎯→ c C + d D a A + b B ⎯⎯ k −1 [C]𝑐 [D]𝑑 Law of mass action 𝐾𝑐 = [A]𝑎 [B]𝑏 The equilibrium-constant expression depends only on the stoichiometry of the reaction, not on its mechanism. Copyright © 2020 Pearson Canada Inc. 14 - 10 Exercise Writing Equilibrium-Constant Expressions Copyright © 2020 Pearson Canada Inc. 14 - 11 Gas Phase In the gas phase, it is more convenient to write the equilibrium constant in terms of pressures. Copyright © 2020 Pearson Canada Inc. 14 - 12 Example: Calculating KP for a gas phase reaction At 248 K and at equilibrium, the partial pressures of N2O5, NO2 and O2 were found to be 3.8, 2.7 and 0.67 atm, respectively. Determine KP? 2 𝑁2 𝑂5 (𝑔) ֎ 4𝑁𝑂2 (𝑔) + 𝑂2 𝑔 (𝑃𝑁𝑂2 )4 𝑃𝑂2 𝐾𝑝 = (𝑃𝑁2𝑂5 )2 Copyright © 2020 Pearson Canada Inc. 14 - 13 Relating KP and KC 𝑎A(𝑔) + 𝑏B(𝑔) ֎ 𝑐C(𝑔) + 𝑑D(𝑔) 𝑐 𝑑 𝑐 𝑐 𝑑 𝑑 𝑃C 𝑃D (𝑐+𝑑)−(𝑎+𝑏) [C][C][D] [D] 𝑅𝑇 𝑅𝑇 𝑃𝐶𝑐 𝑃𝐷𝑑 1 𝐾𝑐𝐾= = c [A]𝑎 [B] 𝑏 𝑏 = 𝑎 𝑏 = 𝑎 𝑏 [A]𝑎 [B] 𝑃A 𝑃B 𝑃𝐴 𝑃𝐵 𝑅𝑇 𝑅𝑇 𝑅𝑇 (𝑐+𝑑)−(𝑎+𝑏) 1 𝐾P = 𝐾c (𝑅𝑇)(𝑐+𝑑)−(𝑎+𝑏) 𝐾c = 𝐾P 𝑅𝑇 𝐾P = 𝐾c (𝑅𝑇)Δn Copyright © 2020 Pearson Canada Inc. 14 - 14 Exercise Converting between Kc and Kp 𝐾P = 𝐾c (𝑅𝑇)Δn Copyright © 2020 Pearson Canada Inc. 14 - 15  Equilibrium Constants are Unitless k ⎯⎯→ a A + b B ⎯⎯1 c C+d D 𝑎 A(g) + 𝑏 B(g) ֎ 𝑐 C(g) + 𝑑 D(g) k −1 [C]𝑐 [D]𝑑 𝑃𝐶𝑐 𝑃𝐷𝑑 𝐾𝑐 = 𝐾𝑝 = 𝑎 𝑏 [A]𝑎 [B]𝑏 𝑃𝐴 𝑃𝐵 Units of Kc and Kp would seem depend on the stoichiometry of the reaction 𝑢𝑛𝑖𝑡 𝑜𝑓 𝐾𝑐 = 𝑀 𝑐+𝑑 −(𝑎+𝑏) 𝑢𝑛𝑖𝑡 𝑜𝑓 𝐾𝑝 = 𝑎𝑡𝑚( 𝑐+𝑑 − 𝑎+𝑏 ) But K are unitless numbers. Copyright © 2020 Pearson Canada Inc. 14 - 16 Equilibrium Constants are Unitless Referenced to the standard state A concentration of 1 mol L−1 for a solute (c° = 1 mol L−1) A pressure of 1 atm for a gas (P° = 1 atm) Assuming the aqueous or gaseous solutions to be ideal, the activities for a component, X, are defined as: aX = [X]/c° for a solute 𝑋 𝑚𝑜𝑙Τ𝐿/ 1 𝑚𝑜𝑙Τ𝐿 aX = PX/P° for a gas 𝑃 𝑎𝑡𝑚Τ1 𝑎𝑡𝑚 Fore a pure liquid aliquid = 1 For a pure solid asolid =1 Pure solids and liquids need not be included in the equilibrium- constant expression. Copyright © 2020 Pearson Canada Inc. 14 - 17 Equilibrium Can Be Reached from Either Direction The ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 were. Copyright © 2020 Pearson Canada Inc. 14 - 18 Achieving Equilibrium: The Haber Process Consider the Haber process, which is the industrial preparation of ammonia: N2 𝑔 + 3 H2 𝑔 ⇌ 2 NH3 𝑔 The equilibrium constant depends on stoichiometry: NH3 2 𝐾𝑐 = N2 H2 3 Copyright © 2020 Pearson Canada Inc. 14 - 19 Achieving Equilibrium The figure shows that it doesn’t matter whether you start with reactants or products; an equilibrium will be reached. N2 + 3 H2 ⇌ 2 NH3 Relative amount of change is based on stoichiometry. Copyright © 2020 Pearson Canada Inc. 14 - 20 Achieving Equilibrium N2 + 3 H2 ⇌ 2 NH3 It doesn’t matter whether we start with N2 and H2 or whether we start with NH3. We will have the same proportions of all three substances at once equilibrium is reached. Copyright © 2020 Pearson Canada Inc. 14 - 21 Using Equilibrium Constants Magnitude of K If K>>1, the reaction favors products. Products predominates at equilibrium. If K