Chemistry

4,211 views
3,846 views

Published on

Published in: Technology, Business
0 Comments
3 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
4,211
On SlideShare
0
From Embeds
0
Number of Embeds
5
Actions
Shares
0
Downloads
95
Comments
0
Likes
3
Embeds 0
No embeds

No notes for slide
  • Answer: A
  • Answer: A
  • Answer: A
  • Answer: A
  • Answer: C
  • Answer: C
  • Answer: C
  • Answer: C
  • Answer: B
  • Answer: B
  • Answer: B
  • Answer: B
  • Answer: C
  • Answer: C
  • Chemistry

    1. 1. CHAPTER 5 ThermochemistryCHEMISTRYTHIRD EDITIONGilbert | Kirss | Foster | Davies © 2012 by W. W. Norton & Company
    2. 2. Chapter Outline 5.1 Energy: Basic Concepts and Definitions » Work, Potential Energy, and Kinetic Energy » Kinetic Energy and Potential Energy at the Molecular Level 5.2 Systems, Surroundings, and Energy Transfer 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity 5.5 Calorimetry: Measuring Heat Capacity and Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values 5.8 Hess’s Law © 2012 by W. W. Norton & Company
    3. 3. Definitions Thermodynamics: • Transformation of energy from one form to another. Thermochemistry: • Energy in the form of heat consumed or produced by chemical reactions. • 2H2(g) + O2(g) → 2H2O(l) + energy Energy: Heat vs Work © 2012 by W. W. Norton & Company
    4. 4. Definitions (cont.) Heat: • Energy transferred between objects because of a difference in their temperatures. Work: w = F × d • Work (w) is done when a force (F) moves an object through a distance (d). © 2012 by W. W. Norton & Company
    5. 5. Work and Energy © 2012 by W. W. Norton & Company
    6. 6. Two Types of Energy Potential: due to position or composition: • Can be converted to work: PE = m × g × h » m = mass, g = force of gravity, and h = vertical distance • Chemical energy = a form of potential energy Kinetic: due to motion of the object • KE = 1/2 mu2 (m = mass, u = velocity) © 2012 by W. W. Norton & Company
    7. 7. Potential Energy: A State Function Depends only on the difference between initial and final state of the system. • Independent of path between states. © 2012 by W. W. Norton & Company
    8. 8. The Nature of Energy Law of Conservation of Energy: • Energy can be neither created nor destroyed. • Can be converted from one form to another. » Potential energy → kinetic energy » Chemical energy → heat © 2012 by W. W. Norton & Company
    9. 9. Energy at the Molecular Level Kinetic energy at the molecular level depends on: • Mass and velocity of the particle (KE = ½ mu2). • Temperature » As T increases, molecular motion and KE increase. Potential energy at molecular level: • Electrostatic attractions: Q1 × Q2 Eel ∝ d Where Eel is electrostatic energy, Q1 and Q2 are charges separated by distance, d. © 2012 by W. W. Norton & Company
    10. 10. Electrostatic Potential Energy © 2012 by W. W. Norton & Company
    11. 11. Chapter Outline 5.1 Energy: Basic Concepts and Definitions 5.2 Systems, Surroundings, and Energy Transfer » Isolated, Closed, and Open Systems » Exothermic and Endothermic Processes » Energy Units and P-V Work 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity 5.5 Calorimetry: Measuring Heat Capacity and Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values 5.8 Hess’s Law © 2012 by W. W. Norton & Company
    12. 12. Terms Describing Energy Transfer System: The part of the universe that is the focus of a thermodynamic study. • Isolated / open / closed Surroundings: Everything in the universe that is not part of the system. Universe = System + Surroundings © 2012 by W. W. Norton & Company
    13. 13. Examples © 2012 by W. W. Norton & Company
    14. 14. Heat Flow Exothermic process: Heat flows out of system to surroundings (q < 0). Endothermic process: Heat flows into system from surroundings (q > 0). © 2012 by W. W. Norton & Company
    15. 15. Phase Changes and Heat FlowEndothermic Exothermic © 2012 by W. W. Norton & Company
    16. 16. Energy and Phase Changes Absorbed heat increases kinetic energy of molecules. Loss of kinetic energy caused by release of heat by molecules. © 2012 by W. W. Norton & Company
    17. 17. Internal Energy Internal energy of a system = sum of all KE and PE of all components of the system. • Different types of molecular motion contribute to overall internal energy: (a) translational, (b) rotational, and (c) vibrational. © 2012 by W. W. Norton & Company
    18. 18. First Law of Thermodynamics First Law of Thermodynamics = Law of Conservation of Energy! Energy of the universe is constant! • Universe = system + surroundings, so… • energy gained or lost by a system must equal the energy lost or gained by surroundings. © 2012 by W. W. Norton & Company
    19. 19. Units of Energy Calorie (cal): • The amount of heat necessary to raise the temperature of 1 g of water 1°C. Joule (J): • The SI unit of energy. • 4.184 J = 1 cal. Energy = heat and/or work (same units!). © 2012 by W. W. Norton & Company
    20. 20. Energy Flow Diagrams © 2012 by W. W. Norton & Company
    21. 21. Change in Internal Energy ΔE = q + w: • ΔE = change in system’s internal energy • q = heat • w = work Work: • w = −PΔV » where P = pressure, ΔV = change in volume. • Work done by the system is energy lost by the system (added to the surroundings), hence the negative sign. © 2012 by W. W. Norton & Company
    22. 22. Problem: Calculation of WorkCalculate the work in L∙atm and joulesassociated with the expansion of a gas in acylinder from 54 L to 72 L at a constant externalpressure of 18 atm. (Note: 1 L∙atm = 101.32 J) • Collect and Organize: • Analyze: • Solve: • Think about It: © 2012 by W. W. Norton & Company
    23. 23. Chapter Outline 5.1 Energy: Basic Concepts and Definitions 5.2 Systems, Surroundings, and Energy Transfer 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity 5.5 Calorimetry: Measuring Heat Capacity and Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values 5.8 Hess’s Law © 2012 by W. W. Norton & Company
    24. 24. Enthalpy and Change in Enthalpy Enthalpy (H) = E + PV Change in Enthalpy (ΔH) = ΔE + PΔV ΔH = change in enthalpy; energy flows as heat at constant pressure: ΔH = qP = ΔE + PΔV ΔH > 0, Endothermic; ΔH < 0, Exothermic Add subscripts to indicate ΔH for specific process or part of the universe; e.g., ΔHvap, ΔHrxn, ΔHsys. © 2012 by W. W. Norton & Company
    25. 25. Chapter Outline 5.1 Energy: Basic Concepts and Definitions 5.2 Systems, Surroundings, and Energy Transfer 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity » How does heat flow translate into changes in temperature or phase changes? 5.5 Calorimetry: Measuring Heat Capacity and Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values 5.8 Hess’s Law © 2012 by W. W. Norton & Company
    26. 26. Heating Curves Heat in → kinetic energy © 2012 by W. W. Norton & Company
    27. 27. Heating Curves Heat in → phase change © 2012 by W. W. Norton & Company
    28. 28. Heat Capacities Molar heat capacity (cp) is the heat required to raise the temperature of 1 mole of a substance by 1°C at constant pressure. • q = ncpΔT (cp = J / (mol∙°C) Specific heat (cs) is the heat required to raise the temperature of 1 gram of a substance by 1°C at constant pressure. • q = ncsΔT (cs = J / (g∙°C) Heat capacity (Cp) is the quantity of heat needed to raise the temperature of some specific object by 1°C at constant pressure. © 2012 by W. W. Norton & Company
    29. 29. Molar Heat Capacities © 2012 by W. W. Norton & Company
    30. 30. Heating Curves heat in, q = ncp(g)ΔT heat in, q = ncp(l)ΔT heat in, q = ncp(s)ΔT © 2012 by W. W. Norton & Company
    31. 31. Calculating Energy Through a Change in State Molar heat of fusion: • ΔHfus = heat needed to convert 1 mole of a solid at its melting point to 1 mole of liquid. • q = nΔHfus Molar heat of vaporization: • ΔHvap = heat needed to convert 1 mole of a liquid at its boiling point to 1 mole of vapor. • q = nΔHvap © 2012 by W. W. Norton & Company
    32. 32. Heating Curves heat in, q = nΔHvap heat in, q = nΔHfus © 2012 by W. W. Norton & Company
    33. 33. Problem: Specific Heat CapacityDuring a strenuous workout, a student generates 2000kJ of heat energy. What mass of water would have toevaporate from the student’s skin to dissipate thismuch heat? • Collect and Organize: • Analyze: • Solve: • Think about It: © 2012 by W. W. Norton & Company
    34. 34. Problem: Heat Gain vs Heat LossExactly 10 mL of water at 25°C was added to a hotiron skillet. All of the water was converted into steam at100°C. If the mass of the pan was 1.20 kg and themolar heat capacity of iron is 25.19 J/mol∙°C, what wasthe temperature change of the skillet? • Collect and Organize: • Analyze: • Solve: • Think about It: © 2012 by W. W. Norton & Company
    35. 35. Chapter Outline 5.1 Energy: Basic Concepts and Definitions 5.2 Systems, Surroundings, and Energy Transfer 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity 5.5 Calorimetry: Measuring Heat Capacity and Calorimeter Constants » Determining Molar Heat Capacity and Specific Heat » Measuring Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values 5.8 Hess’s Law © 2012 by W. W. Norton & Company
    36. 36. Calorimetry Calorimetry = measurement of heat. • Typically, measuring change in heat that accompanies a physical change or chemical process. A calorimeter = device used to measure the absorption or release of heat by a physical or chemical process. • A closed system! −qsystem = qcalorimeter © 2012 by W. W. Norton & Company
    37. 37. Measuring Heat Capacity(a) 23.5 g of Al beads are heated to 100.0°C in boiling water. (heat transferred from water to Al until the final temp of Al beads = 100.0°C) © 2012 by W. W. Norton & Company
    38. 38. Measuring Heat Capacity (cont.)(b) The Al beads (at 100°C) are placed ina Styrofoam box calorimeter containing 103.0 gof water at 23.0°C. (c) Heat is transferred from Al beads to water until they reach a thermal equilibrium at 26.0°C; heat lost by Al = heat gained by H2O. −qaluminum = qwater mAlcs(Al)ΔTal = mwatercs(water)ΔTwater © 2012 by W. W. Norton & Company
    39. 39. Measuring Heats of Reaction (ΔHrxn) A bomb calorimeter is a constant-volume device used to measure the heat of a combustion reaction. Heat produced by reaction = heat gained by calorimeter ΔH = −qcal = −CcalΔT © 2012 by W. W. Norton & Company
    40. 40. Chapter Outline 5.1 Energy: Basic Concepts and Definitions 5.2 Systems, Surroundings, and Energy Transfer 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity 5.5 Calorimetry: Measuring Heat Capacity and Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction » Standard Enthalpies of Formation » A Practical Application 5.7 Fuel Values and Food Values 5.8 Hess’s Law © 2012 by W. W. Norton & Company
    41. 41. Heat of Reaction Heat of reaction: • Also known as enthalpy of reaction (ΔHrxn). • The heat absorbed or released by a chemical reaction. Table 5.2B: Some Standard Enthalpies of Combustion (ΔHcomb°) © 2012 by W. W. Norton & Company
    42. 42. A Specific Enthalpy The standard enthalpy of formation, ΔHf°: • The enthalpy change of the formation reaction. • A formation reaction is the process of forming 1 mole of a substance in its standard state from its component elements in their standard states. • For example, formation reaction for water: » H2(g) + ½ O2(g) → H2O(l) » ΔHrxn = ΔHf°(H2O) (The standard state of a substance is its most stable form under 1 bar pressure and at 25°C.) © 2012 by W. W. Norton & Company
    43. 43. Methods of Determining ΔHrxn1. From calorimetry experiments: • ΔHrxn = −Ccal ΔT2. From enthalpies of formation: • ΔHrxn° = ΣnpΔHf°(products) − ΣnrΔHf°(reactants) • ΔHf° values for substances in Appendix 4.§ Using Hess’s Law (Section 5.8). © 2012 by W. W. Norton & Company
    44. 44. Problem: Using ΔHf° to Find ΔHrxn°Use Table 5.2 to calculate an approximateenthalpy of reaction for CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). • Collect and Organize: • Analyze: • Solve: • Think about It: © 2012 by W. W. Norton & Company
    45. 45. Problem 2: Using ΔHf° to Find ΔHrxn°One step in the production of nitric acid is thecombustion of ammonia. Use data in Appendix4 to calculate the enthalpy of this reaction: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) • Collect and Organize: • Analyze: • Solve: • Think about It: © 2012 by W. W. Norton & Company
    46. 46. Chapter Outline 5.1 Energy: Basic Concepts and Definitions 5.2 Systems, Surroundings, and Energy Transfer 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity 5.5 Calorimetry: Measuring Heat Capacity and Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values » Fuel Values vs. Fuel Densities » Food Value 5.8 Hess’s Law © 2012 by W. W. Norton & Company
    47. 47. Fuel Values Amount of energy (in kJ/g) from combustion reaction. • CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) • ΔHcomb = 802.3 kJ/mol • Fuel value = (802.3 kJ/mol)∙(1 mol/16.04 g) = 50.02 kJ/g Fuel Density: • For liquid fuels, energy released in kJ/L. © 2012 by W. W. Norton & Company
    48. 48. Fuel Values (cont.) Molecular Fuel Value Compound Formula (kJ/g)*Methane CH4 50.0Ethane C2H6 47.6Propane C3H8 46.3Butane C4H10 45.8* Based on the formation of H2O (g) © 2012 by W. W. Norton & Company
    49. 49. Food Values Amount of energy produced when food is burned completely: • Determined by bomb calorimetry. • Nutritional Calorie = 1 kcal = 4.184 kJ. Food Food Value Food Value Category (Cal or kcal) (kJ) Proteins 4.0 16.7 Carbohydrates 4.0 16.7 Fats 9.0 37.7 © 2012 by W. W. Norton & Company
    50. 50. Chapter Outline 5.1 Energy: Basic Concepts and Definitions 5.2 Systems, Surroundings, and Energy Transfer 5.3 Enthalpy and Enthalpy Changes 5.4 Heating Curves and Heat Capacity 5.5 Calorimetry: Measuring Heat Capacity and Calorimeter Constants 5.6 Enthalpies of Formation/Enthalpies of Reaction 5.7 Fuel Values and Food Values 5.8 Hess’s Law » Calculating ΔHrxn by Combining Reactions © 2012 by W. W. Norton & Company
    51. 51. Hess’s Law Hess’s Law of constant heat of summation: • The ΔH of a reaction that is the sum of two or more reactions is equal to the sum of the ΔH values of the constituent reactions.1. CH4(g) + H2O(g) → CO(g) + 3H2(g) ΔH12. CO(g) + 3H2(g) + H2O(g) → 4H2(g) + CO2(g) ΔH23. CH4(g) + 2H2O(g) → 4H2(g) + CO2(g) ΔH3 ΔH3 = ΔH1 + ΔH2 © 2012 by W. W. Norton & Company
    52. 52. Hess’s Law (cont.) © 2012 by W. W. Norton & Company
    53. 53. Calculations Using Hess’s Law1. If a reaction is reversed, ΔH sign changes. N2(g) + O2(g) → 2NO(g) ΔH = 180 kJ 2NO(g) → N2(g) + O2(g) ΔH = −180 kJ2. If the coefficients of a reaction are multiplied by an integer, ΔH is multiplied by that same integer. 6NO(g) → 3N2(g) + 3O2(g) ΔH = 3(−180 kJ) ΔH = −540 kJ © 2012 by W. W. Norton & Company
    54. 54. Problem: Using Hess’s LawUsing the following data, calculate the enthalpy change forthe reaction: C2H4(g) + H2(g) → C2H6(g) ΔHrxnH2(g) + ½O2(g) → H2O(l) −285.8 kJC2H4(g) + 3O2(g) → 2H2O(l) + 2CO2(g) −1411 kJC2H6(g) + 7/2O2(g) → 3H2O(l) + 2CO2(g) −1560 kJ • Collect and Organize: • Analyze: • Solve: • Think about It: © 2012 by W. W. Norton & Company
    55. 55. ChemTour: State Functions and Path Functions Click here to launch this ChemTourThis ChemTour defines and explores the difference between state andpath functions using a travel analogy that leads into a discussion ofenergy, enthalpy, heat, and work. © 2012 by W. W. Norton & Company
    56. 56. ChemTour: Internal Energy Click here to launch this ChemTourThis ChemTour explores how energy is exchanged between a system and itssurroundings as heat and/or work, and how this transfer in turn affects theinternal energy (E) of a system. It includes Practice Exercises. © 2012 by W. W. Norton & Company
    57. 57. ChemTour: Pressure–Volume Work Click here to launch this ChemTourAn animated ChemTour of an internal combustion engine shows how a system undergoing anexothermic reaction can do work on its surroundings; students can explore the relationshipamong pressure, volume, and work. It includes Practice Exercises. © 2012 by W. W. Norton & Company
    58. 58. ChemTour: Heating Curves Click here to launch this ChemTourIn this ChemTour, students use interactive heating curve diagrams to explore phase changes,heat of fusion, and heat of vaporization. Macroscopic views of ice melting and water boiling areshown in sync with the appropriate sections of the heating curve. It includes PracticeExercises. © 2012 by W. W. Norton & Company
    59. 59. ChemTour: Calorimetry Click here to launch this ChemTourThis ChemTour demonstrates how a bomb calorimeter works andwalks students through the equations used to solve calorimetryproblems. It includes an interactive experiment. © 2012 by W. W. Norton & Company
    60. 60. ChemTour: Hess’s Law Click here to launch this ChemTourThis ChemTour explains Hess’s law of constant heat of summationusing animated sample problems and step-by-step descriptions. Itincludes Practice Exercises. © 2012 by W. W. Norton & Company
    61. 61. Sample Exercise 5.1Describe the flow of heatduring the purification ofwater by distillation(Figure 5.13), identify thesteps in the process aseither endothermic orexothermic, and give thesign of q associated witheach step. Consider thewater being purified to bethe system. © 2012 by W. W. Norton & Company
    62. 62. Sample Exercise 5.1 (cont.) Collect and Organize: Since the water is the system, we must evaluate how the water gains or loses energy during distillation. Analyze: In distillation, energy flows in three steps: (1) liquid water is heated to the boiling point and (2) vaporizes. (3) The vapors are cooled and condense as they pass through the condenser. © 2012 by W. W. Norton & Company
    63. 63. Sample Exercise 5.1 (cont.) Solve: a. Energy flows from the surroundings (hot plate) to heat the impure water (the system) to its boiling point and then to vaporize it. Therefore, processes 1 and 2 are endothermic. The sign of q is positive for both. b. Because energy flows from the system (water vapor) into the surroundings (condenser walls), process 3 is exothermic. Therefore, the sign of q is negative. © 2012 by W. W. Norton & Company
    64. 64. Sample Exercise 5.1 (cont.) Think about It: “Endothermic” means that energy is transferred from the surroundings into the system, the water in the distillation flask. When the water vapor is cooled in the condenser, energy flows from the vapor as it is converted from a gas to a liquid; the process is exothermic. © 2012 by W. W. Norton & Company
    65. 65. Sample Exercise 5.2 Figure 5.19 shows a simplified version of a piston and cylinder in an engine. Suppose combustion of fuel injected into the cylinder produces 155 J of energy. The hot gases in the cylinder expand, pushing the piston down. In doing so, the gases do 93 J of P–V work on the piston. If the system is the gases in the cylinder, what is the change in internal energy of the system? Collect and Organize: The change in internal energy is related to the work done by a system or on a system and the heat gained or lost by the system (Equation 5.5). First we have to decide whether q and w are positive or negative according to the sign convention shown in Figure 5.18. © 2012 by W. W. Norton & Company
    66. 66. Sample Exercise 5.2 (cont.) Analyze: The system (gases in the cylinder) absorbs energy, so q > 0, and the system does work on the surroundings (the piston), so w < 0. Solve: ΔE = q + w = (155 J) + (-93 J) = 62 J Think about It: More energy enters the system (155 J) than leaves it (93 J), so a positive value of ΔE is reasonable. © 2012 by W. W. Norton & Company
    67. 67. Sample Exercise 5.3A tank of compressed helium is used to inflateballoons for sale at a carnival on a day when theatmospheric pressure is 1.01 atm. If eachballoon is inflated from an initial volume of 0.0 Lto a final volume of 4.8 L, how much P–V work isdone by 100 balloons on the surroundingatmosphere when they are inflated? Theatmospheric pressure remains constant duringthe filling process. © 2012 by W. W. Norton & Company
    68. 68. Sample Exercise 5.3 (cont.) Collect and Organize: Each of 100 balloons goes from empty (V = 0.0 L) to 4.8 L, which means ΔV = 4.8 L, and the atmospheric pressure P is constant at 1.01 atm. The identity of the gas used to fill the balloons doesn’t matter, because under normal conditions all gases behave the same way, regardless of their identity. Our task is to determine how much P–V work is done by the 100 balloons on the air that surrounds them. © 2012 by W. W. Norton & Company
    69. 69. Sample Exercise 5.3 (cont.) Analyze: We focus on the work done on the atmosphere (the surroundings) by the system; the balloons and the helium they contain are the system. Solve: The volume change (ΔV) as all the balloons are inflated is 100 balloons × 4.8 L/balloon = 480 L The work (w) done by our system (the balloons) as they inflate against an external pressure of 1.01 atm is Because the work is done by the system on its surroundings, the work is negative from the point of view of the system: © 2012 by W. W. Norton & Company
    70. 70. Sample Exercise 5.3 (cont.) Think about It: The internal energy of the balloons (the system) decreases when they are inflated. Air injected into a balloon cools when the balloon expands and does work on the surroundings. The work is done by the system. © 2012 by W. W. Norton & Company
    71. 71. Sample Exercise 5.4 Between periods of a hockey game, an ice-refinishing 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 ΔHsys as the water freezes? b. To freeze this volume of water at 0°C, what is the value of ΔHsys? The density of water is 1.00 g/mL. Collect and Organize: The water from the ice-refinishing machine is identified as the system, so we can determine the sign of ΔH by determining whether heat transfer is into or out of the water. We are told the volume of the water and its density, so we can calculate the mass of water involved in the change of state from liquid water to ice. © 2012 by W. W. Norton & Company
    72. 72. Sample Exercise 5.4 (cont.) Analyze: (a) Because the freezing takes place at constant pressure, ΔHsys = 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, ΔHsolid = -6.01 kJ/mol. © 2012 by W. W. Norton & Company
    73. 73. Sample Exercise 5.4 (cont.) Solve: a. For the water (the system) to freeze, energy must be removed from it. Therefore, ΔHsys must be a negative value. b. We convert the volume of water into moles: and then calculate ΔHsys: © 2012 by W. W. Norton & Company
    74. 74. Sample Exercise 5.4 (cont.) Think about It: ΔH changes with pressure but the magnitude of the change is negligible for pressures near normal atmospheric pressure. The magnitude of the answer to part b is reasonable given the large amount of water that is frozen to refinish the rink’s ice. © 2012 by W. W. Norton & Company
    75. 75. Sample Exercise 5.5 Calculate the amount of energy required to raise the temperature of 237 g of solid ice from 0.0°C to 80.0°C. The molar heat of fusion (ΔHfus) of ice is 6.01 kJ/mol. The molar heat capacity of liquid water is 75.3 J/(mol . °C). Collect and Organize: This problem refers to a process symbolized by segments and in Figure 5.21. The ice must first be melted, and we are given the molar heat of fusion of ice. The amount of energy required to heat the water formed when all the ice has been melted can be determined using the molar heat capacity of liquid water, and Equation 5.8. We can calculate the temperature change (ΔT) of the water from the initial and final temperatures. © 2012 by W. W. Norton & Company
    76. 76. Sample Exercise 5.5 (cont.) Analyze: Four mathematical steps are required: (1) calculate the number of moles of ice; (2) determine the amount of energy required to melt the ice (3) calculate the amount of energy required to raise the liquid water temperature from 0.0°C to 80.0°C; (4) add the results of steps (2) and (3). We can calculate the number of moles in 237 g of water using the molar mass of water (M =18.02 g/mol). The water increases in temperature from 0.0°C to 80.0°C. Solve: 1. Calculate the number of moles of water: © 2012 by W. W. Norton & Company
    77. 77. Sample Exercise 5.5 (cont.)2. Determine the amount of energy needed to melt theice (Equation 5.9):3. Determine the amount of energy needed to warm thewater (Equation 5.8):4. Add the results of parts 2 and 3: © 2012 by W. W. Norton & Company
    78. 78. Sample Exercise 5.5 (cont.) Think about It: Using the definitions of molar heat of fusion and molar heat capacity, we can solve this exercise without referring back to Equations 5.8 and 5.9. Molar heat of fusion defines the amount of energy needed to melt 1 mole of ice. Multiplying that value (6.01 kJ/mol) by the number of moles (13.2 mol) gives us the energy required to melt the given amount of ice. By the same token, the molar heat capacity of water [75.3 J/(mol . °C )] defines the amount of energy needed to raise the temperature of 1 mole of liquid water by 1 °C.We know the number of moles (13.2 mol), and we know the number of degrees by which we want to raise the temperature (80.0 °C - 0.00°C = 80.0°C ); multiplying those factors together gives us the heat needed to raise the temperature of the water. © 2012 by W. W. Norton & Company
    79. 79. Sample Exercise 5.6 If you add 250.0 g of ice initially at -18.0°C to 237 g (1 cup) of freshly brewed tea initially at 100.0°C and the ice melts, what is the final temperature of the tea? Assume that the mixture is an isolated system (in an ideal insulated container) and that tea has the same molar heat capacity, density, and molar mass as water. Collect and Organize: We know the mass of tea, its initial temperature, and the molar heat capacity of tea, for which we just use the cP value for water. The amount of energy released when the tea is cooled will be the same as the amount of energy gained by the ice and, once it is melted, the amount gained by the water that is formed as it warms to the final temperature. We know the amount of ice, its initial temperature, and the heat of fusion of ice. Our task is to find the final temperature of the tea ice–water mixture. © 2012 by W. W. Norton & Company
    80. 80. Sample Exercise 5.6 (cont.) Analyze: Before solving this problem, we need to think about the changes that take place when the tea and ice come into contact. Assuming the ice melts completely, three heat transfers account for the energy lost by the tea: 1. q1: raising the temperature of the ice to 0.0°C 2. q2: melting the ice. 3. q3: bringing the mixture to the final temperature; the temperature of the water from the melted ice rises and the temperature of the tea (Tinitial = 100.0°C ) falls to the final temperature of the mixture (Tfinal = ?). The energy gained by the ice equals the energy lost by the tea: qice = -qtea Based on our analysis, we know that qice + q1 + q2 + q3 © 2012 by W. W. Norton & Company
    81. 81. Sample Exercise 5.6 (cont.) Solve: The energy lost by the tea as it cools from 100.0°C to Tfinal is, from Equation 5.8, The transfer of this heat is responsible for the changes in the ice. We can treat the heat transfer from the hot tea to the ice in terms of the three processes we identified in Analyze. In step 1, the ice is warmed from -18.0°C (Tinitial) to 0.0°C (Tfinal in this step): © 2012 by W. W. Norton & Company
    82. 82. Sample Exercise 5.6 (cont.)In step 2, the ice melts, requiring the absorption ofenergy:In step 3, the water from the ice, initially at 0.0°C, warmsto the final temperature (where ΔT = Tfinal - Tinitial = Tfinal -0.0°C ): © 2012 by W. W. Norton & Company
    83. 83. Sample Exercise 5.6 (cont.)The sum of the quantities of energy absorbed by the iceand the water from it during steps 1 through 3 mustbalance the energy lost by the tea:Expressing all values in kilojoules: © 2012 by W. W. Norton & Company
    84. 84. Sample Exercise 5.6 (cont.) and rearranging the terms to solve for Tfinal, we have Think about It: This calculation was carried out assuming the system (the tea plus the ice) to be isolated and the vessel to be a perfect insulator. Our answer, therefore, is an “ideal” answer and reflects the coldest temperature we can expect the tea to reach. In the real world, the ice would absorb some energy from the surroundings (the container and the air) and the tea would lose some energy to the surroundings, so the final temperature of the beverage could be different from the ideal value we calculated. © 2012 by W. W. Norton & Company
    85. 85. Sample Exercise 5.7 Before we can determine the enthalpy change of a reaction run in a calorimeter, we must determine the heat capacity (the calorimeter constant) of the calorimeter. What is the calorimeter constant of a bomb calorimeter if burning 1.000 g of benzoic acid in it causes the temperature of the calorimeter to rise by 7.248°C ? The heat of combustion of benzoic acid is ΔHcomb = -26.38 kJ/g. Collect and Organize: We are asked to find the calorimeter constant of a calorimeter. We are given data describing how much the temperature of the calorimeter rises when a known amount of benzoic acid is burned in it, and we are given the heat of combustion of benzoic acid. © 2012 by W. W. Norton & Company
    86. 86. Sample Exercise 5.7 (cont.) Analyze: We need to determine the amount of energy required to raise the temperature of the calorimeter by 1°C. The heat capacity of a calorimeter can be calculated using Equation 5.13 and the knowledge that the combustion of 1.000 g of benzoic acid produces 26.38 kJ of heat. © 2012 by W. W. Norton & Company
    87. 87. Sample Exercise 5.7 (cont.) Solve: This calorimeter can now be used to determine the heat of combustion of any combustible material. © 2012 by W. W. Norton & Company
    88. 88. Sample Exercise 5.7 (cont.) Think about It: The calorimeter constant is determined for a specific calorimeter. If anything changes—if the thermometer breaks and has to be replaced, or if the calorimeter loses any of the water it contains—a new constant must be determined. © 2012 by W. W. Norton & Company
    89. 89. Sample Exercise 5.8Which of the following reactions are formationreactions at 25°C? For those that are not, explainwhy not. © 2012 by W. W. Norton & Company
    90. 90. Sample Exercise 5.8 (cont.) Collect and Organize: We are given four balanced chemical equations and information about the standard states of specific reactants and products. Analyze: For a reaction to be a formation reaction, it must meet the criteria that it produces 1 mole of a substance from its component elements in their standard states. Each reaction must therefore be evaluated for the quantity of product and for the state of each reactant and product. © 2012 by W. W. Norton & Company
    91. 91. Sample Exercise 5.8 (cont.) Solve: a. The reaction shows 1 mole of water vapor formed from its constituent elements in their standard states. This is the formation reaction for H2O(g), and its enthalpy of reaction is correctly symbolized ΔH°f. b. The reaction shows 1 mole of liquids methanol formed from its constituent elements in their standard states. This reaction therefore is a formation reaction, and its change in enthalpy is correctly symbolized ΔH°f. © 2012 by W. W. Norton & Company
    92. 92. Sample Exercise 5.8 (cont.)c. This is not a formation reaction because the reactants arenot elements in their standard states and because more thanone product is formed. The change in enthalpy of this reactionwould be symbolized ΔHcomb.d. This is not a formation reaction because the product is 4moles of POCl3. Note, however, that all of the constituentelements are in their standard states and that only onecompound is formed. We could therefore call the heat ofreaction (ΔHrxn) for this reaction a standard heat of reaction,which we indicate by adding a superscript degree to thesymbol: ΔH°rxn. We could easily convert this into a formationreaction by dividing all the coefficients and the standard heatof reaction (ΔH°rxn) by 4. Once we do this, ΔH°rxn = ΔH°f. © 2012 by W. W. Norton & Company
    93. 93. Sample Exercise 5.8 (cont.) Think about It: Just because we can write formation reactions for substances like methanol does not mean that anyone would ever use that reaction to make methanol. Remember that formation reactions are defined to provide a standard against which other reactions can be compared when their thermochemistry is evaluated. © 2012 by W. W. Norton & Company
    94. 94. Sample Exercise 5.9 Using the appropriate values from Table 5.2, calculate ΔH°rxn for the combustion of the fuel propane (C3H8) in air. Collect and Organize: The reactants are propane and elemental oxygen, O2, and the products are carbon dioxide and water. Even though combustion is an exothermic reaction, we assume that water is produced as liquid, i.e., H2O(ℓ), so that all products and reactants are in their standard states. The heats of formation of propane, carbon dioxide, and H2O(ℓ) are given in Table 5.2. The heat offormation of the element O 2 in its standard state is zero. Our task is to use the balanced equation and the data from Table 5.2 to calculate the heat of combustion. © 2012 by W. W. Norton & Company
    95. 95. Sample Exercise 5.9 (cont.) Analyze: Equation 5.17 defines the relation between heats of formation of reactants and products and the standard enthalpy of reaction. We also need the balanced equation for combustion of propane: © 2012 by W. W. Norton & Company
    96. 96. Sample Exercise 5.9 (cont.) Solve: Inserting ΔH°f values for the products [CO2(g) and H2O(ℓ)] and reactants [C3H8(g) and O2(g)] from Table 5.2 and the coefficients in the balanced chemical equation into Equation 5.17, we get © 2012 by W. W. Norton & Company
    97. 97. Sample Exercise 5.9 (cont.) Think about It: The result of the calculation has a large negative value, which means that the combustion reaction is highly exothermic, as expected for a hydrocarbon fuel. © 2012 by W. W. Norton & Company
    98. 98. Sample Exercise 5.10Most automobiles run on either gasoline ordiesel fuel. Although both fuels are mixtures, theenergy content in gasoline can be approximatedby considering it to have the formula C9H20 (d =0.718 g/mL), while diesel fuel may beconsidered to be C14H30 (d = 0.763 g/mL). Usingthese two formulas, compare (a) fuel value pergram of each fuel and (b) fuel density per liter ofeach fuel. © 2012 by W. W. Norton & Company
    99. 99. Sample Exercise 5.10 (cont.) Collect and Organize: We are given representative chemical formulas for gasoline (C9H20) and diesel fuel (C14H30) and can calculate the molar mass for each fuel. Heat of combustion data for C9H20 and C14H30 are given in Table 5.2. Our task is to use these data and the respective molar masses to obtain the fuel values. We can then use the given density of each fuel to convert the fuel value from kJ/g to kJ/L. © 2012 by W. W. Norton & Company
    100. 100. Sample Exercise 5.10 (cont.) Analyze: The enthalpies of combustion in Table 5.2 are given in kilojoules per mole, so to answer this question in terms of grams and liters, we need molar masses to convert moles to grams in part a and for part b we need density (d = m/V, grams per milliliter) to convert grams to liters. Taking C9H20 and C14H30 as the average molecules in regular gasoline and diesel fuel, respectively, we can determine their molar masses. Then the fuel value can be calculated from the heats of combustion. Using the densities given in the problem, we can calculate fuel densities in kJ/L. © 2012 by W. W. Norton & Company
    101. 101. Sample Exercise 5.10 (cont.) Solve: a. Fuel values b. Fuel densities © 2012 by W. W. Norton & Company
    102. 102. Sample Exercise 5.10 (cont.) Think about It: We wouldn’t expect the fuel values of gasoline and diesel fuel to be very different, or otherwise there would be a strong preference for gasoline-fueled cars over diesel or vice versa. The values are similar to the fuel values of methane and propane calculated in the text, so these answers seem reasonable. In the United States, fuel for cars and trucks is bought by the gallon and its consumption is rated in miles per gallon, whereas in most countries it is bought by the liter (1 U.S. gal = 3.785 L). Because of the way we buy automobile fuels, it makes sense to compare fuel densities rather than fuel values. On either basis, diesel is a slightly inferior fuel compared to gasoline. © 2012 by W. W. Norton & Company
    103. 103. Sample Exercise 5.11 Glucose (C6H12O6) is a simple sugar formed by photosynthesis in plants. The complete combustion of 0.5763 g of glucose in a calorimeter (Ccalorimeter = 6.20 kJ/°C ) raises the temperature of the calorimeter by 1.45°C . What is the food value of glucose in Calories per gram? Collect and Organize: We are asked to determine the food value of glucose, which means the energy given off when 1 g is burned. We have the mass of glucose burned and the calorimeter constant. We can relate the energy given off by the glucose to the energy gained by the calorimeter (Equation 5.15) and the energy given off by the glucose to the temperature change and the calorimeter constant (Equation 5.12). © 2012 by W. W. Norton & Company
    104. 104. Sample Exercise 5.11 (cont.) Analyze: We use the data from the calorimetry experiment to determine how much heat in kilojoules is given off when the stated amount of glucose is burned. We can convert that quantity into Calories by using the conversion factor 1 Cal = 4.184 kJ. © 2012 by W. W. Norton & Company
    105. 105. Sample Exercise 5.11 (cont.) Solve: To convert this quantity of energy to a food value, we divide by the sample mass: and then convert into Calories: © 2012 by W. W. Norton & Company
    106. 106. Sample Exercise 5.11 (cont.) Think about It: One gram of a doughnut, which is mostly carbohydrates, and 1 g of glucose have about the same food value (19 kJ/g and 15.6 kJ/g, respectively), so the answer to this Sample Exercise seems reasonable. © 2012 by W. W. Norton & Company
    107. 107. Sample Exercise 5.12Hydrocarbons burned in a limited supply of air maynot burn completely, and CO(g) may be generated.One reason furnaces and hot-water heaters fueledby natural gas need to be vented is that incompletecombustion can produce toxic carbon monoxide: © 2012 by W. W. Norton & Company
    108. 108. Sample Exercise 5.12 (cont.) Collect and Organize: We are given two reactions (B and C) with thermochemical data and a third (A) for which we are asked to find ΔH °comb. All the reactants and products of reaction A are present in reactions B and/or C. Analyze: We can manipulate the equations for reactions B and C so that they sum to give the equation for which ΔH °comb is unknown. Then we can calculate this unknown value by applying Hess’s law. The reaction of interest (A) has methane on the reactant side. Because reaction B also has methane as a reactant, we can use B as written. Reaction A has CO as a product. Reaction C involves CO as a reactant, so we have to reverse C in order to get CO on the product side. Once we reverse C, we must change the sign of ΔH°comb. If the coefficients as given do not allow us to sum the two reactions to yield reaction A, we can multiply one or both reactions by other factors. © 2012 by W. W. Norton & Company
    109. 109. Sample Exercise 5.12 (cont.) Solve: We start with reaction B as written and add the reverse of reaction C, remembering to change the sign of (C) ΔH°comb: Because methane has a coefficient of 2 in reaction A, we multiply all the terms in reaction B, including ΔH°comb (B), by 2: © 2012 by W. W. Norton & Company
    110. 110. Sample Exercise 5.12 (cont.)Because the carbon monoxide in reaction A has acoefficient of 2, we do not need to multiply reactionC by any factor. Now we add (2 × B) to the reverseof C and cancel out common terms: © 2012 by W. W. Norton & Company
    111. 111. Sample Exercise 5.12 (cont.) Think about It: We used Hess’s law to calculate the heat of combustion of methane to make CO, which is impossible to achieve in an experiment because any CO produced can react with O2 to give CO2, which results in a mixture of CO and CO2 as a product. The answer of -1038 kJ is a little less negative than the heat of combustion for the production of CO2 from methane, so the answer is reasonable. © 2012 by W. W. Norton & Company
    112. 112. Clicker Questions Note Note to InstructorsThe following in-class questionsare also applicable to Chapter 12,The Chemistry of Solids. © 2012 by W. W. Norton & Company
    113. 113. Clicker Question: Isothermal Expansion of Ideal Gas An ideal gas in a sealed piston is allowed to expand isothermally (at a constant temperature) against a pressure of 1 atm. In what direction, if at all, does heat flow for this process? (Hint: E = q − PΔV)A) into the system B) out of the system C) heat does not flow © 2012 by W. W. Norton & Company
    114. 114. Clicker Question: Isothermal Expansion of Ideal GasConsider the following arguments for each answer and vote:A. When the gas expands isothermally, it does work without a decrease in its energy, so heat must flow into the system.B. During the expansion, the gas pressure decreases, thereby releasing heat to the surroundings.C. The fact that the process is isothermal means that heat does not flow. © 2012 by W. W. Norton & Company
    115. 115. Clicker Question: Adiabatic Compression of an Ideal Gas An ideal gas in an insulated piston is compressed adiabatically (q = 0) by its surroundings. What can be said of the change in the temperature (ΔT) of the gas for this process?A) ΔT > 0 B) ΔT = 0 C) ΔT < 0 © 2012 by W. W. Norton & Company
    116. 116. Clicker Question: Adiabatic Compression of an Ideal GasConsider the following arguments for each answer and vote:A. The surroundings are doing work on the system, and no heat is flowing. Therefore, ΔE > 0 and ΔT > 0.B. The volume of the gas decreases, but the pressure increases to keep the product of the pressure and volume constant. Therefore, the temperature is also constant.C. The gas is being compressed to a more ordered state, which corresponds to a lower temperature. © 2012 by W. W. Norton & Company
    117. 117. Clicker Question:ΔT of a Released Rubber Band Consider a stretched rubber band that is suddenly released. What can be said of the change in the temperature (ΔT) of the rubber band for this process?A) ΔT > 0 B) ΔT = 0 C) ΔT < 0 © 2012 by W. W. Norton & Company
    118. 118. Clicker Question: ΔT of a Released Rubber BandConsider the following arguments for each answer and vote:A. The stretched rubber band is at a higher energy state than the unstretched rubber band. Releasing the stretched rubber band causes the energy to be released.B. Because the recoil of the rubber band is rapid, this process is essentially adiabatic. Therefore, the temperature of the rubber band will not change.C. As the rubber band contracts, it does work and its energy decreases, resulting in a decrease in its temperature. © 2012 by W. W. Norton & Company
    119. 119. Clicker Question: Specific Heat Capacity of Al and FeA 1.0 gram block of Al (cs = 0.9 J∙°C − 1∙g − 1) at 100°C and a 1.0 gram block of Fe (cs = 0.4 J∙°C − 1∙g − 1) at 0 °C are added to 10 mL of water (cs = 4.2 J∙°C − 1∙g − 1) at 50°C. What will be the final temperature of the water? A) < 50°C B) 50°C C) > 50°C © 2012 by W. W. Norton & Company
    120. 120. Clicker Question: Specific Heat Capacity of Al and FeConsider the following arguments for each answer and vote:A. The specific heat capacity of Fe(s) is smaller than that of Al(s), so heat from both the Al(s) and the water will be required to warm the Fe(s).B. The average initial temperature of the three components is 50°C. Therefore, the final temperature of the water will be 50°C.C. The specific heat capacity of Al(s) is greater than that of Fe(s), so the Al block at 100°C will heat the water more than the Fe block will cool it. © 2012 by W. W. Norton & Company
    121. 121. Clicker Question: Equilibrium Constant of CyclooctatetraeneCyclooctatetraene, C8H8, can undergo a transformation between two possible states, A and B, by rearranging its 4 double bonds.Which of the following graphs depicts the dependenceof the equilibrium constant (K) on temperature for theconversion from state A to state B? A) B) C) © 2012 by W. W. Norton & Company
    122. 122. Clicker Question: Equilibrium Constant of CyclooctatetraeneConsider the following arguments for each answer and vote:A. The intermediate is at a higher energy state than either states A or B, so at high temperatures, the reaction will favor the intermediate and K will decrease.B. The enthalpies of formation for states A and B are equal, so ΔH° = 0 and K is not temperature dependent.C. At high temperatures, the conversion from state A to state B will occur at a much faster rate, thus increasing the value of K. © 2012 by W. W. Norton & Company
    123. 123. Clicker Question:Enthalpies of N2, NH3, and N2O The reaction of N2(g) and O2(g) to form N2O(g) is an endothermic process. The reaction of N2(g) and H2(g) to form NH3(g) is an exothermic process. Given this information, which of the following species has the lowest enthalpy of formation?A) N2(g) B) NH3(g) C) N2O(g) © 2012 by W. W. Norton & Company
    124. 124. Clicker Question: Enthalpies of N2, NH3, and N2OConsider the following arguments for each answer and vote:A. N2(g) is the elemental form of nitrogen, which by definition will have ΔHf° = 0, the lowest possible enthalpy of formation.B. Since the formation reaction for N2O(g) is endothermic, it has a higher ΔHf° than N2(g), O2(g), and H2(g); NH3(g) has a negative ΔHf°, less than N2(g), O2(g), and H2(g).C. The formation of an N—N double bond and an N—O double bond, as found in N2O, releases more energy than does the creation of 3 N—H bonds to form NH3(g). © 2012 by W. W. Norton & Company
    125. 125. Clicker Question: Polymerization of EthyleneWhich of the following is true of ΔH° for the polymerization of ethylene to form polyethylene? Note: the C—C single bond enthalpy is ~350 kJ/mole and the C—C double bond enthalpy is ~600 kJ/mole. A) ΔH° > 0 B) ΔH° = 0 C) ΔH° < 0 © 2012 by W. W. Norton & Company
    126. 126. Clicker Question: Polymerization of EthyleneConsider the following arguments for each answer and vote:A. The energy required to break double bonds is more than the energy released by forming new single bonds.B. The total number of C—C bonds (if we count double bonds twice) does not change with polymerization. Therefore, there can be no change in ΔH°.C. For each C-C double bond that breaks (~600 kJ/mole), two single bonds form (2 × ~350 = ~700 kJ/mole). © 2012 by W. W. Norton & Company

    ×