Chapter 6 Thermochemistry Lecture Slides PDF

Chapter 6 Thermochemistry: Energy Changes in Reactions Copyright © 2020 W. W. Norton & Company Chapter Outline 6.1 Sunlight Unwinding 6.2 Forms of Energy 6.3 Systems, Surroundings, and Energy Transfer 6.4 Enthalpy and Enthalpy Change 6.5 Heating Curves, Molar Heat Capacity, and Specific Heat 6...

Chapter 6 Thermochemistry: Energy Changes in Reactions Copyright © 2020 W. W. Norton & Company Chapter Outline 6.1 Sunlight Unwinding 6.2 Forms of Energy 6.3 Systems, Surroundings, and Energy Transfer 6.4 Enthalpy and Enthalpy Change 6.5 Heating Curves, Molar Heat Capacity, and Specific Heat 6.6 Calorimetry: Measuring Heat Capacity and Enthalpies of Reaction 6.7 Hess’s Law 6.8 Standard Enthalpies of Formation and Reaction 6.9 Fuel, Fuel Values, and Food Values 2 Learning Outcomes Determine the flow of energy (q) and enthalpy change associated with a change of state (fusion, vaporization, etc). or with changing the temperature of a substance Measure and calculate enthalpies of reaction 3 Thermodynamics Thermodynamics – the study of energy and its transformations Thermochemistry – the study of the relation between chemical reactions and changes in energy Thermochemical equation: 2 H2(g) + O2(g) → 2 H2O(ℓ) + energy Thermal equilibrium – a condition in which temperature is uniform throughout a material and no energy flows from one point to another 4 Heat and Work Heat − the energy transferred between objects because of a difference in their temperatures Work − a form of energy: the energy required to move an object through a given distance w=F×d w is work. F is force. d is distance. 5 Two Types of Energy Potential energy (PE) – the energy stored in an object because of its position PE = m × g × h m = mass g = acceleration due to gravity h = vertical distance Kinetic energy (KE) – the energy due to motion of the object KE = 1/2mu2 m = mass u = velocity 6 Work and Energy 7 Potential Energy: A State Function Depends only on the difference between initial and final state of the system and not how it achieved that state 8  Total Energy Total energy = PE + KE = m·g·h + ½mu2 9 The Nature of Energy Law of conservation of energy Energy can be neither created nor destroyed. Energy can be converted from one form to another. Potential energy → kinetic energy Chemical energy → heat Thermal energy – kinetic energy of atoms, ions, and molecules 10 Energy at the Molecular Level Kinetic energy at the molecular level: Mass, velocity of the particle (KE = ½mu2) Temperature As T increases, molecular motion and KE increase. Potential energy at the molecular level: Electrostatic interactions: (𝑄1 × 𝑄2 ) 𝐸𝑒𝑙 ∝ 𝑑 11 Electrostatic Potential Energy 12 Energy of Chemical Reactions 13 Terminology of Energy Transfer System – the part of the universe that is the focus of a thermochemical study Isolated – exchanges no energy or matter with surroundings Closed – exchanges energy but no matter with surroundings Open – exchanges both energy and matter with surroundings Surroundings – everything in the universe that is not part of the system Universe = system + surroundings 14 Examples of Systems 15 Heat Flow Exothermic process – energy flows out of system to surroundings (q < 0). Endothermic process – energy flows into system from surroundings (q > 0). q is quantity of energy transferring. 16 Phase Changes and Heat Flow Energy and Phase Changes Absorbed heat increases kinetic energy of molecules. Loss of kinetic energy is caused by release of heat by molecules. 18 Internal Energy Internal energy (E) State function ∆E = Efinal – Einitial Sum of KE and PE of all components of the system Types of molecular motion (a) translational (b) rotational (c) vibrational 19 Change in Internal Energy ∆E = q + w ∆E = change in system’s internal energy q = heat, w = work Work w = –P∆V where P = pressure, V = change in volume Work done by the system = energy lost by the system, hence the negative sign. ∆E = q + w = q + (–P∆V) = q – P∆V 20 Units of Energy Calorie (cal) The amount of heat necessary to raise the temperature of 1 g of water by 1oC Joule (J) The SI unit of energy 4.184 J = 1 cal Energy = heat and/or work (same units!) 21 First Law of Thermodynamics First law of thermodynamics = law of conservation of energy Energy of the universe is constant! Universe = system + surroundings Energy gained or lost by a system must equal the energy lost or gained by the surroundings. ∆Esystem = –∆Esurroundings 22 Energy Flow Diagram 23 Practice: Work Calculation Collect and Organize Calculate the work in L atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm. (Note: 1 L atm = 101.32 J) We are given the expansion of gas in a cylinder and the pressure. Vinitial = 54 L Vfinal = 72 L P = 18 atm We need to calculate the amount of work performed. Practice: Work Calculation Analyze Calculate the work in L atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm. (Note: 1 L atm = 101.32 J) w = –PV The sign associated with w will be negative, since work is done by the system. Pressure–volume work will have units of L · atm, which we can convert to energy units using the conversion factor 1 L · atm = 101.32 J. Practice: Work Calculation Solve Calculate the work in L atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm. (Note: 1 L atm = 101.32 J) w = –(18 atm) × (72 – 54 L) = –324 L ∙ atm w = –(324 L ∙ atm) × (101.32 J/L ∙ atm) = –32827 J = –33 kJ Practice: Work Calculation Think About It Calculate the work in L atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm. (Note: 1 L atm = 101.32 J) The negative sign indicates work done by the system—energy lost from the system to the surroundings. Practice: Work Calculation Summary Collect and Organize: We are given the expansion of gas in a cylinder and the pressure. We need to calculate the amount of work performed. Vinitial = 54 L Vfinal = 72 L P = 18 atm Analyze: The sign associated with w will be negative, since work is done by the system. Pressure–volume work will have units of L·atm, which we can convert to energy units (joules) using the conversion factor (1 L · atm = 101.32 Solve: w = –(18 atm) × (72 – 54 L) = –324 L ∙ atm w = –(324 L ∙ atm) × (101.32 J/L ∙ atm) = –32827 J = –33 kJ Think About It: The negative sign indicates work done by the system—energy lost from the system to the surroundings. Enthalpy, Change in Enthalpy Enthalpy (H): ∆H = ∆E + P∆V H = E + PV ∆H = qP = ∆E + P∆V Enthalpy change (∆H): ∆H = ∆E + P∆V ∆H > 0, endothermic; ∆H < 0, exothermic Subscripts indicate ∆H for specific processes. 29 Enthalpy Change 30 Practice: Determining the Value and Sign of ∆H Between periods of a hockey game, an ice-resurfacing machine spreads 855 L of water across the surface of a hockey rink. (a) If the system is the water, what is the sign of ∆Hsys as the water freezes? (b) To freeze this volume of water at 0°C, what is the value of ∆Hsys? The density of water is 1.00 g/mL. Practice: Determining the Value and Sign of ∆H Collect and Organize Between periods of a hockey game, an ice-resurfacing machine spreads 855 L of water across the surface of a hockey rink. (a) If the system is the water, what is the sign of ∆Hsys as the water freezes? (b) To freeze this volume of water at 0°C, what is the value of ∆Hsys? The density of water is 1.00 g/mL. We are given the volume and density of the water, and we need to determine the sign and value of ∆Hsys. The water from the ice-resurfacing machine is the system, so we can determine the sign of ∆H by determining whether energy transfer is into or out of the water. Practice: Determining the Value and Sign of ∆H Analyze Between periods of a hockey game, an ice-resurfacing machine spreads 855 L of water across the surface of a hockey rink. (a) If the system is the water, what is the sign of ∆Hsys as the water freezes? (b) To freeze this volume of water at 0°C, what is the value of ∆Hsys? The density of water is 1.00 g/mL. (a) The freezing takes place at constant pressure, so ∆Hsys = qP. (b) To calculate the amount of energy lost from the water as it freezes, we must convert 855 L of water into moles of water because the conversion factor between the quantity of energy removed and the quantity of water that freezes is the enthalpy of solidification of water, ∆Hsolid = 6.01 kJ/mol. Practice: Determining the Value and Sign of ∆H Solve Between periods of a hockey game, an ice-resurfacing machine spreads 855 L of water across the surface of a hockey rink. (a) If the system is the water, what is the sign of ∆Hsys as the water freezes? (b) To freeze this volume of water at 0°C, what is the value of ∆Hsys? The density of water is 1.00 g/mL. (a) For the water to freeze, energy must be removed from it. Therefore, ∆Hsys must be a negative value. (b) Convert the volume of water into moles. 1000 mL 1.00 g 1 mol 855 L × × × = 4.745 × 104 mol 1L 1 mL 18.02 g −6.01 kJ Then, calculate ∆Hsys: 4.745 × 104 mol × 1 mol = −2.85 × 105 kJ Practice: Determining the Value and Sign of ∆H Summary Collect and Organize: We are given the volume and density of the water, and we need to determine the sign and value of ∆Hsys. Analyze: (a) The freezing takes place at constant pressure, so ∆Hsys = qP. (b) To calculate the amount of energy lost from the water as it freezes, we must convert 855 L of water into moles of water because the conversion factor between the quantity of energy removed and the quantity of water that freezes is the enthalpy of solidification of water, ∆Hsolid = 6.01 kJ/mol. Solve: 1000 mL 1.00 g 1 mol 855 L × × × = 4.745 × 104 mol 1L 1 mL 18.02 g 4 −6.01 kj 4.745 × 10 mol × = −2.85 × 105 kj 1 mol Think About It: The magnitude of the answer to part (b) is reasonable given the large amount of water that is frozen to refinish the rink’s surface. Heating Curves Heat in → kinetic energy 36 Heat Capacities Molar heat capacity (cP) Quantity of energy required to raise the temperature of 1 mole of a substance by 1℃ q = ncPT cP = J/(mol ℃) Specific heat (cs) The energy required to raise the temperature of 1 gram of a substance by 1oC q = ncsT cs = J/(g ℃) Heat capacity (CP) Quantity of energy needed to raise the temperature of an object by 1℃ 37 Enthalpy in Change of State Molar heat of fusion (Hfus) The energy required to convert 1 mole of a solid at its melting point into the liquid state q = nHfus Molar heat of vaporization (Hvap) The energy required to convert 1 mole of a liquid at its boiling point to the vapor state q = nHvap 38 Phase Changes in Heating Curve 39 Cooling Curves Heat transfer: ice @ –8.0C to water @ 0.0C 1. Temp change: q1 = ncPT 2. Phase change: q2 = nHfus 40 Calorimetry Calorimetry The measurement of quantity of heat transferred during a physical or chemical process Calorimeter A device used to measure the absorption or release of heat by a physical or chemical process Closed system: –qsystem = qcalorimeter 41 Calorimetry: Hrxn HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(ℓ) 42 Practice: Calorimetry Collect and Organize 25.0 mL of 1.0 M HCl in a coffee cup calorimeter has a temperature of 18.5°C. Rapid addition of 25.0 mL of 1.0 M NaOH causes the temperature to rise to 25.0°C. What is the ∆Hrxn? Cp water = 75.3 J/mol oC We know the balanced equation is HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(ℓ) We have the volume and concentration of HCl and NaOH. We also have the initial and final temperatures. 25.0 mL of 1.0 M HCl 25.0 mL of 1.0 M NaOH Initial temp = 18.5oC Final temp = 25.0oC Practice: Calorimetry Analyze 25.0 mL of 1.0 M HCl in a coffee cup calorimeter has a temperature of 18.5°C. Rapid addition of 25.0 mL of 1.0 M NaOH causes the temperature to rise to 25.0°C. What is the ∆Hrxn? Cp water = 75.3 J/mol oC We are producing water so we can focus on the amount of energy in the water. Assume the density of water = 1.0 g/mL. q rxn 𝑞H2 O = 𝑛H2 O 𝐶𝑝 ∆𝑇 ∆𝐻𝑟𝑥𝑛 = mol rxn Practice: Calorimetry Solve 25.0 mL of 1.0 M HCl in a coffee cup calorimeter has a temperature of 18.5°C. Rapid addition of 25.0 mL of 1.0 M NaOH causes the temperature to rise to 25.0°C. What is the ∆Hrxn? Cp water = 75.3 J/mol oC 1.00 g H2 O 1.00 mol H2 O 𝑛H2 O = 50.0 mL H2 O × × = 2.77 mol H2 O 1 mL H2 O 18.02 g H2 O 75.3 J 𝑞H2 O = 2.77 mol H2 O × × 25.0°C − 18.5℃ = 1356 J mol ∙ °C q H2 O = −q rxn = −1356 J −1356 J ∆𝐻rxn = = −54kJ/mol 0.025 mol Practice: Calorimetry Summary Collect and Organize: We know the balanced equation and we have the volume and concentration of HCl and NaOH. We also have the initial and final temperatures. Analyze: We are producing water so we can focus on the amount of energy in the water. Assume the density of water = 1.0 g/mL. 1.00 g H2 O 1.00 mol H2 O Solve: 𝑛H2 O = 50.0 mL H2 O × 1 mL H2 O × 18.02 g H2 O = 2.77 mol H2 O 75.3 J 𝑞H2 O = 2.77 mol H2 O × × 25.0°C − 18.5℃ = 1356 J mol ∙ °C qH2 O = −qrxn = −1356 J −1356 J ∆𝐻rxn = = −54kJ/mol 0.025 mol Think About It: The solution is going up in temperature so the reaction is releasing energy and so the ∆Hrxn should be negative. Heats of Reaction Bomb calorimeter A constant-volume device used to measure the energy released during a combustion reaction Heat produced by reaction = heat gained by calorimeter qcal = CcalT = –Hrxn 47 Hess’s Law Hess’s law of constant heat of summation The Hrxn for a reaction that is the sum of two or more reactions is equal to the sum of the Hrxn values of the constituent reactions. 1. CH4(g) + H2O(g) → CO(g) + 3 H2(g) H1 2. CO(g) + 3 H2(g) + H2O(g) → 4 H2(g) + CO2(g) H2 3. CH4(g) + 2 H2O(g) → 4 H2(g) + CO2(g) H3 H3 = H1 + H2 48 Calculations Using Hess’s Law N2(g) + O2(g) → 2 NO(g) H = 180 kJ 1. If a reaction is reversed, the sign of H changes. 2 NO(g) → N2(g) + O2(g) H = –180 kJ 2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by the same integer. 6 NO(g) → 3 N2(g) + 3 O2(g) H = 3(–180 kJ) H = –540 kJ 49  Practice: Hess’s Law Collect and Organize Calculate the Hrxn for the reaction using the following reactions and ∆Hrxn values: C2H4(g) + H2(g) → C2H6(g) H2(g) + ½ O2(g) → H2O(ℓ) –285.8 kJ C2H4(g) + 3 O2(g) → 2 H2O(ℓ) + 2 CO2(g) –1411 kJ C2H6(g) + (7/2) O2(g) → 3 H2O(ℓ) + 2 CO2(g) –1560 kJ We have H values for three reactions and need to combine them in such a way that we obtain the reaction: C2H4 + H2 → C2H6  Practice: Hess’s Law Analyze Calculate the Hrxn for the reaction using the following reactions and ∆Hrxn values: C2H4(g) + H2(g) → C2H6(g) H2(g) + ½ O2(g) → H2O(ℓ) –285.8 kJ C2H4(g) + 3 O2(g) → 2 H2O(ℓ) + 2 CO2(g) –1411 kJ C2H6(g) + (7/2) O2(g) → 3 H2O(ℓ) + 2 CO2(g) –1560 kJ The first two reactions have the reactants (C2H4 and H2) on the left side, while the third reaction has the desired product on the left side. We will need to reverse the third reaction to get the desired product on the right side: 1. H2(g) + ½ O2(g) → H2O(ℓ); H1 = –285.8 kJ 2. C2H4(g) + 3 O2(g) → 2 H2O(ℓ) + 2 CO2(g); H2 = –1411 kJ 3. 3 H2O(ℓ) + 2 CO2(g) → 7/2 O2(g) + C2H6(g); H3 = –(–1560 kJ)  Practice: Hess’s Law Solve Calculate the Hrxn for the reaction using the following reactions and ∆Hrxn values: C2H4(g) + H2(g) → C2H6(g) H2(g) + ½ O2(g) → H2O(ℓ) –285.8 kJ C2H4(g) + 3 O2(g) → 2 H2O(ℓ) + 2 CO2(g) –1411 kJ C2H6(g) + (7/2) O2(g) → 3 H2O(ℓ) + 2 CO2(g) –1560 kJ We now add these three reactions together. H2O and O2 will cancel on the left and right sides, yielding the desired reaction. So, the Hrxn can be calculated as: Hrxn = H1 + H2 – H3 Hrxn = –285.8 kJ + (–1441 kJ) + 1560 kJ = –136.8 KJ Hrxn = –137 kJ Practice: Hess’s Law Summary Collect and Organize: We have H values for three reactions and need to combine them in such a way that we obtain the reaction: C2H4 + H2 → C2H6 Analyze: The first two reactions have the reactants (C2H4 and H2) on the left side, while the third reaction has the desired product on the left side. We will need to reverse the third reaction to get the desired product on the right side. Solve: We now add these three reactions together. H2O and O2 will cancel on the left and right sides, yielding the desired reaction. So, the Hrxn can be calculated as: Hrxn = H1 + H2 – H3 Hrxn = –285.8 kJ + (–1441 kJ) + 1560 kJ = –136.8 KJ Hrxn = –137 kJ Think About It: Since the first two Hrxn values were negative it would make sense that the answer be a negative value. Enthalpy of Formation, Hof Standard enthalpy of formation, Hf Enthalpy change that takes place at constant pressure when 1 mole of a substance is formed from its elements in their standard state Formation reaction for water H2(g) + ½ O2(g) → H2O(ℓ) Hrxn = Hof,H2O Standard state of a substance: is its most stable form under standard conditions Standard conditions: 1 bar pressure at a given temperature, usually 25C 54 Standard Enthalpy of Reaction Standard enthalpy of reaction (Hrxn) Energy associated with a reaction that takes place under standard conditions Calculated from Hf Hrxn = nproductsHf,products – nreactantsHof,reactants 55 Methods of Determining Hrxn 1. By calorimetry experiments Hrxn = –CcalT 2. By enthalpies of formation Hrxn = nproductsHf,products – nreactantsHf,reactants Hf values for substances are in Table 5.2. 3. Using Hess’s law 56 Alkanes Alkanes Hydrocarbons where each carbon atom is bonded to four other atoms The names for alkanes are derived from prefixes based on the number of carbons. Formula: CnH2n+2 57 Models 58 Melting Points and Boiling Points for Alkanes 59 Straight-Chain and Branched Hydrocarbons 60 Fuel Values (1 of 2) The energy released during complete combustion of 1 g of a substance CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) Hcomb = –802.3 kJ/mol Fuel value = (802.3 kJ/mol)(1 mol/16.04 g) = 50.02 kJ/g Fuel density – the energy released during complete combustion of 1 L of a liquid fuel 61 Fuel Values (2 of 2) 62 Food Values Food value The quantity of energy produced when a material consumed by an organism for sustenance is burned completely Determined by bomb calorimetry Nutritional calorie = 1 kcal = 4.184 kJ 63