2. Thermochemistry Chapter 8

3. Energy Changes in Chemical Reactions 90 0 C Which one has more thermal energy? Coffee cup? Bathtub?

4. Energy Changes in Chemical Reactions 90 0 C 40 0 C Which one has more thermal energy? What additional information is needed? How does it relate to heat?

5. Heat: the transfer of thermal energy between two bodies that are at different temperatures. Energy Changes in Chemical Reactions Temperature is a measure of the thermal energy. Measure of the average kinetic energy Temperature = Thermal Energy

6. Energy: the capacity to do work Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position. E = mgh Kinetic Energy: energy of motion E = ½ mu 2 Energy

7. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing SYSTEM SURROUNDINGS Energy Changes

8. Detected by generation of Heat: energy that flows from a body at higher temperature to a body of lower temperature Represented by the letter q. Work: can be done on the system by surroundings or vice versa. Examples: hammering a nail into a wooden block; gasoline burning in an engine; tossing a baseball (or any other ball) T ∝ KE KE = ½ mv 2 Units of Energy: Joules calories : 1 calorie = 4.18 J Calories = food calorie = 1000 calories

9. Temperature Degree of hotness or coldness. Measures the average KE of the molecules. Thermometers operate on the following principles: 1. expansion of liquids 2. expansion of gas 3. radiation properties of substances 4. electrical properties (digital thermometers) Temperature scales: Kelvin Celsius Fahrenheit

10. The higher the temperature, the greater the average speed of the molecules. T   ½ mv 2 Velocity or speed T  v Temperature

11. Energy Changes State Properties: depend only on the initial and final states of the system as it is defined by volume, temperature, pressure, number of moles.

12. Exothermic process: is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process: is any process in which heat is supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) energy + H 2 O (s) H 2 O (l) Heat: Endothermic/Exothermic

13. Enthalpy: Enthalpy Diagrams ΔH rxn = ΔH products – ΔH reactants

14. Enthalpy Enthalpy : a thermodynamic quantity used to describe heat changes at constant pressure (most reactions occur at constant pressure). Heat absorbed or released at const. P during chemical reaction. Equation: ΔH reaction = ΔH products - ΔH reactants

15. Enthalpy: Examples Indicate the sign of enthalpy change in the following processes carried out under atmospheric pressure, and indicate whether the process is exothermic or endothermic: 1.An ice cube melts 2.1 g of butane (C 4 H 10 ) is combusted in sufficient oxygen to give complete combustion to CO 2 and H 2 O 3.A bowling ball is dropped from a height of 8 ft into a bucket of sand.

16. Thermochemical Equations Find the differences between the following equations: 1.CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) ΔH = -890.4 kJ 2.CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g) ΔH = -802.4 kJ Thermochemical Equations: describe the reaction (includes the state of the substances) and energy changes that occur during the chemical reaction.

17. Thermochemical Equations H 2 O (s) → H 2 O (l)  H = 6.01 kJ Is  H negative or positive? System absorbs heat Endothermic  H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. 6.3

18. Thermochemical Equations CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ Is  H negative or positive? System gives off heat Exothermic  H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm. 6.3

19. Rules of Thermochemistry ΔH directly proportional to the amount of reactants or products (stoichiometric relationship). ΔH forward = - ΔH reverse  The value of ΔH for the reaction is the same whether it occurs in one step or in series of steps. (Hess’ Law- AP Chem)  The physical states of all reactants and products must be specified in a thermochemical equation.

20. Rule #1: ΔH is directly proportional to the amount of reactants or products (stoichiometric relationship). C 6 H 6 (l) → C 6 H 6 (g) ΔH = 7.36 kCal 2C 6 H 6 (l) ΔH = ½ C 6 H 6 (l) ΔH = Given the thermochemical equation C (s) + O 2 (g) → CO 2 (g) ΔH = -94.1 kCal Calculate (a) ΔH when 1.00 g of C burns. (b) ΔH when 1.8 g of CO 2 is produced.

21. Thermochemical Equations: Rules How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s)  H = -3013 kJ 266 g P 4 1 mol P 4 123.9 g P 4 x -3013 kJ 1 mol P 4 x [ -6470 kJ]

22. Example Given the thermochemical equation: C 8 H 18 (l) + 12.5 O 2 (g) → 8CO 2 (g) + 9H 2 O(l) ΔH = -1308 kCal How many grams of octane have to be burned to evolve 1 kilocalorie of heat? Fractions are allowed

23. Thermochemical Equations: Reverse Reaction When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes. H 2 O(s) → H 2 O(l) ΔH = 1.44 kCal CaCO 3 (s) → CaO(s) + CO 2 (g) ΔH = 42.5 kCal C 10 H 8 (s) + 12 O 2 (g) → 10CO 2 (g) + H 2 O(l) ΔH = -1232.5 kCal

24. Thermochemical Equations: Rules The physical states of all reactants and products must be specified in thermochemical equations. H 2 O(s) → H 2 O(l) ΔH = 6.00 kJ H 2 O(l) → H 2 O(g) ΔH = 44.0 kJ What will be the ΔH when 1 mole of ice at 0ºC is changed into one mole of steam at the boiling point (100 0ºC)?

25. Hess’ Law Heat of reaction is an algebraic sum of heats of chemical reactions that when added give the required equation. Allow calculations of ΔH for reactions that cannot be easily experimentally determined. The value of the reaction is the same whether it occurs in one step or in series of steps. Example: H 2 (g) + ½ O 2 (g) → H 2 O(g)ΔH = -57.8 kCal H 2 (g) + ½ O 2 (g) → H 2 O(l)ΔH = -68.3 kcal Needed: H 2 O(g) → H 2 O(l), cannot be easily measured

26. Calorimetry

27. Heat thermal energy transferred from a hot object to a cold object. The heat transferred is proportional to the mass specific heat capacity temperature change Calorimetry Heat has the symbol q and is calculated using … q = mcΔT

28. Calorimetry Quantity of heat depends on: 1. The mass of the substance 2. Mass of the calorimeter 3. Specific heat of the substance and calorimeter. Heat capacity, C: quantity of heat needed to raise the temperature of a given substance by 1ºC. Extensive or Intensive? ___________ Specific heat, c or c p : heat needed to raise the temperature of 1.0 g of substance by 1ºC. Extensive or Intensive? ___________

29. q = m c  T Quantity of heat mass specific heat capacity temperature change Law of Conservation of Energy

30. 1. How much heat is needed to raise the temperature of 25.6 grams of water from 20.0 C to 50.0 C? The specific heat of water is equal to 4.18 J/(g ºC) Answer: 3210 J 2.How much heat is lost by the system when a solid Al ingot with a mass of 4110 g cools from 60.0 ºC to 25.0 ºC? (-2.35 x 10 6 J) 3. What is the final temperature of 27.0 grams of liquid water, initially at 0 ºC, after it absorbs 700.0 J of energy? (6.20 ºC) Examples

31. 4. A 40.0g of glass was heated from 0.0 ºC to 41.0 ºC and was found to have absorbed 32 J of heat. (a) What is the speicfic heat of this type of glass? (b) How much heat did the same glass sample gain when it was heated from 41.0 3. A 40.0g of glass was heated from 41.0 ºC to 70.0 ºC? 0.20 J/(g ºC); 23 J

32. Examples 5.When 2.80 g CaCl2 dissolves in 100.0 g of water, the temperature of the awater risaes from 20.5 ºC to 25.4 ºC. Assume that all the heat is being absorbed by the water (cp = 4.18 J/(g ºC). (a) Write a balance equation of the solution process. (b) What is q for the solution process? (c) Is the process endo or exo? (d) How much heat is absorbed by the water if 1.00 mole of CaCl 2 dissolved?

33. Phase Changes Phase – homogeneous part of system in contact with other parts of system, separated by well-defined boundary e.g., ice in water, subliming dry ice, evaporating isopropanol

34. Phase Changes Hess’ Law for phase changes.

35. Phase Changes and Heating Curves

36. Is heat is absorbed or released during a phase change? How could you measure the heat absorbed or released as substances change phase?

37. Consider ice melting in water. 1.Does the temperature of the water change? 2.Is the water absorbing or releasing heat? 3.Does ice absorb heat or release heat as it melts?

38. Consider ice melting in water. 1.Does the temperature of the water change? 2.Is the water absorbing or releasing heat? 3.Does ice absorb heat or release heat as it melts? No Releasing heat Absorb heat

39. Consider ice melting in water. The word fusion means “melting”. How could you design an experiment to measure the heat of fusion of ice?

40. Phase Changes and Heating Curves

41. Is heat is absorbed or released during a phase change? How could you measure the heat absorbed or released as substances change phase?

42. Consider ice melting in water. 1.Does the temperature of the water change? 2.Is the water absorbing or releasing heat? 3.Does ice absorb heat or release heat as it melts?

43. Consider ice melting in water. 1.Does the temperature of the water change? 2.Is the water absorbing or releasing heat? 3.Does ice absorb heat or release heat as it melts? No Releasing heat Absorb heat

44. Consider ice melting in water. The word fusion means “melting”. How could you design an experiment to measure the heat of fusion of ice?

45. Consider ice melting in water. You could measure the heat lost by some water as it cools. That should equal the heat gained by the ice as it melts. Ice

46. We now know that heat is absorbed or released during a phase change. Heat is absorbed as solids melt, or liquids vaporize.

47. Ice And melts. Heat is absorbed by the ice.

48. … making liquid water One gram of ice at 0C absorbs 334 J as it melts to form water at 0C.

49. water Heat is released by the water as it freezes. 334 joules is released when one gram of water freezes at 0C. Ice

50. Ice absorbs 334 J per gram as it melts at 0C Water releases 334 J per gram as it freezes at 0C

51. Heat is absorbed by the water as it vaporizes. Hotplate

52. 2260 joules is absorbed by one gram of water as it boils at 100C. Hotplate Heat is absorbed by the water as it vaporizes.

53. Hotplate Water absorbs 2260 J/g as it boils at 100 C Steam releases 2260 J/g as it condenses at 100 C

54. Heat is released by water vapor as it condenses. The heat released by condensing water vapor is a major factor in weather phenomena like thunderstorms and hurricanes.

55. 40,000 + feet Thunderhead The heat released by condensing water vapor causes convection and updrafts in thunderstorms.

56. Phase changes occur at a constant temperature as heat is absorbed or released- Why?

57. The heat energy breaks the intermolecular bonds which keep the water in the liquid phase.

58. The heat gained or lost in phase changes can be calculated using … q = mH f q = mH v Heat of fusion (melting) Heat of vaporization

59. The values for water are … H f = 334 J/g H v =2260 J/g Heat of fusion (melting) Heat of vaporization

60. How much heat is absorbed by 150.0 g of ice as it melts at 0 C? q = m H f q = (150.0 g)(334 J/g) q = 50,100 Jor 50.1 kJ

61. Phase changes occur at a single temperature. Water freezes and ice melts at 0C. At sea level, water boils and steam condenses at 100C.

62. Consider the following heating curve for water. 0 100 Time Temp

63. 0 100 Time Temp Ice at –30C absorbs heat. Temperature rises to 0C. Consider the following heating curve for water.

64. 0 100 Time Temp Ice at 0C absorbs heat and melts at constant 0C to make water at 0C. Consider the following heating curve for water.

65. 0 100 Time Temp When all ice melts, water at 0C absorbs heat and temperature rises to 100C. Consider the following heating curve for water.

66. 0 100 Time Temp Water absorbs heat and boils at a constant temperature of 100C. Consider the following heating curve for water.

67. 0 100 Time Temp Temperature of steam rises as it absorbs heat after all of the water boils. Consider the following heating curve for water.

68. What is happening at each segment of the heating curve? 0 100 Time Temp

69. Look at the different regions of the heating curve for water. 0 100 Time Temp Ice Ice and water Water Water and steam Steam Phase changes?

70. The temperature is constant during a phase change. 0 100 Time Temp Ice Ice and water Water Water and steam Steam Phase changes

71. Calculating heat at each segment of the heating curve. 0 100 Time Temp q 1 =mc i  T The temperature of the ice is increasing. The specific heat for ice is 2.05 J/gC.

72. Calculating heat at each segment of the heating curve. 0 100 Time Temp q 2 =mH f q 1 =mc i  T A phase change occurs at a constant temperature. Use the heat of fusion since ice is melting.

73. Calculating heat at each segment of the heating curve. 0 100 Time Temp q 2 =mH f q 1 =mc i  T q 3 =mc w  T The temperature of the water is increasing. The specific heat of water is 4.18 J/gC.

74. Calculating heat at each segment of the heating curve. 0 100 Time Temp q 2 =mH f q 1 =mc i  T q 3 =mc w  T q 4 =mH v A phase change occurs at a constant temperature. Use the heat of vaporization since water is boiling.

75. Calculating heat at each segment of the heating curve. 0 100 Time Temp q 2 =mH f q 1 =mc i  T q 3 =mc w  T q 4 =mH v q 5 =mc s  T The temperature of the steam is increasing. The specific heat of steam is 2.02 J/gC.

76. Calculating heat at each segment of the heating curve. 0 100 Time Temp q 2 =mH f q 1 =mc i  T q 3 =mc w  T q 4 =mH v q 5 =mc s  T Use q=mc  T when there is a temperature change.

77. Calculating heat at each segment of the heating curve. 0 100 Time Temp q 2 =mH f q 1 =mc i  T q 3 =mc w  T q 4 =mH v q 5 =mc s  T Use q=mH f or q=mH v when there is a phase change.

78. Calculating heat at each segment of the heating curve. 0 100 Time Temp q 2 =mH f q 1 =mc i  T q 3 =mc w  T q 4 =mH v q 5 =mc s  T The total amount of heat absorbed is the sum: q tot = q 1 +q 2 +q 3 +q 4 +q 5

79. What would the cooling curve of steam look like? 0 100 Time Temp

80. What would the cooling curve of steam look like? 0 100 Time Temp Ice Ice and water Water Water and steam Steam Heat energy is released at each step.

81. What would the cooling curve of steam look like? 0 100 Time Temp Ice Ice and water Water Water and steam Steam

82. What would the cooling curve of steam look like? 0 100 Time Temp Ice Ice and water Water Water and steam Steam Heat energy is released at each step.

83. There is something else special about a mixture of ice and water. Suppose ice and water were placed into a perfectly insulated container. The mixture would stay at a constant zero degrees Celsius by establishing an equilibrium.

84. An ice/water equilibrium occurs when the rate at which water freezes is equal to the rate at which ice melts. The amount of ice and water will never change. If the container is completely insulated.

85. Ice and water in an insulated container. Acme Digital Thermometer 0.0 C

86. Some ice melts and forms liquid water. Acme Digital Thermometer 0.0 C

87. Some water freezes and forms ice. Acme Digital Thermometer 0.0 C

88. The amounts of ice and water will remain constant… Acme Digital Thermometer 0.0 C

89. …and the mixture of ice and water will remain at a constant 0C. Acme Digital Thermometer 0.0 C

90. Energy Changes Associated with Changes of State: Heating Curve The heat added to the system at the melting and boiling points goes into pulling the molecules farther apart from each other. The temperature of the substance does not rise during the phase change. Compare slopes: s (solid > liquid > gas)

91. How much heat is required to convert 866 g of ice at -10 ° C to steam at 126 ° C? (specific heats: ice = 2.03 J/g ° C, water = 4.18 J/g ° C, steam = 1.99 J/g ° C;  H fus = 6.01 kJ/mol;  H vap = 40.79 kJ/mol) 0oC0oC 100 o C q 1 = q 2 = q 3 = q 4 = q 5 = -10 o C 126 o C  heats of segments = total heat

92. How much heat is required to convert 866 g of ice at -10 ° C to steam at 126 ° C? (specific heats: ice = 2.03 J/g ° C, water = 4.18 J/g ° C, steam = 1.99 J/g ° C;  H fus = 6.01 kJ/mol;  H vap = 40.79 kJ/mol) 0oC0oC 100 o C q 1 = ms 1  t q 2 = n  H fus q 3 = ms 2  t q 4 = n  H vap q 5 = ms 3  t -10 o C 126 o C  heats of segments = total heat

93. 92 Phase Changes Example: How much heat is released by 200 g of H 2 O as it cools from 85.0 o C to 40.0 o C? The specific heat of water is 4.184 J/g o C. (37.6 kJ) Example: The molar heat capacity of ethyl alcohol, C 2 H 5 OH, is 113 J/mol o C. How much heat is required to raise the temperature of 125 g of ethyl alcohol from 20.0 o C to 30.0 o C? 1 mol C 2 H 5 OH = 46.0 g (3.07 kJ)

94. 93 Heating Curve Diagrams: Examples Example: How many joules of energy must be absorbed by 500 g of H 2 O at 50.0 o C to convert it to steam at 120 o C? The molar heat of vaporization of water is 40.7 kJ/mol and the molar heat capacities of liquid water and steam are 75.3 J/mol o C and 36.4 J/mol o C, respectively. (1.26 x 10 3 kJ)

95. Phase Changes Calculate the enthalpy change upon converting 1.00 mole of ice at -25 ºC to water vapor at 125 ºC under constant pressure of 1 atm. Given: Ice c p = 2.00 J/(g K) H 2 O(l), c p = 4.18 J/g K H 2 O(g), c p = 1.84 J/g K ΔH fusion = 6.01 kJ/mol ΔH vaporization = 40.67 kJ/mol Answer: 56.0 kJ

96. 95 Phase Changes Calculate the amount of heat required to convert 150.0 g of ice at -10.0 o C to water at 40.0 o C. specific heat of ice is 2.09 J/g o C Specific heat of water 4.18 J/g o C Heat of fusion of ice 334 J/g (7.83 x 10 4 kJ)

98. H 2 O (s) H 2 O (l)  H = 6.01 kJ Stoichiometric coefficients: equal the number of moles of a substance Thermochemical Equations: Rules If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = - 6.01 kJ If you multiply both sides of the equation by a factor n, then  H must change by the same factor n. 2H 2 O (s) 2H 2 O (l)  H = 12.0 kJ 2 x 6.01kJ/mol =12.0 KJ/2 mol