1. Chapter 51 Thermochemistry

2. 2 The Nature of Energy From Physics: Force – a kind of push or pull on an object. Energy – the capacity to do work. Work – force applied over a distance w = F  d Heat – energy transferred from a warmer object to a cooler object.

3. Chapter 53 The Nature of Energy Kinetic Energy (Thermal Energy) – energy due to motion. Kinetic and Potential Energy

4. Chapter 54 The Nature of Energy Potential Energy (Stored Energy) – the energy an object possesses due to its position. -Potential energy can be converted into kinetic energy. Example: a ball of clay dropping off a building. Kinetic and Potential Energy

5. Chapter 55 The Nature of Energy SI Unit for energy is the joule, J: 1 J = 1kg(m 2 )/s 2 A more traditional unit is the Calorie Calorie (cal) – amount of energy required to raise 1.0 g of water 1 o C. 1cal = 4.184J Energy Units

6. Chapter 56 The Nature of Energy System – portion of the universe we wish to study. Surroundings – everything else. Universe = System + Surroundings Systems and Surroundings

7. Chapter 57 The Nature of Energy System – portion of the universe we wish to study. Surroundings – everything else. Universe = System + Surroundings Systems and Surroundings

8. Chapter 58 First Law of Thermodynamics “The total amount of energy in the universe is fixed.” Also referred to as the “Law of Conservation of Energy”

9. Chapter 59 First Law of Thermodynamics Internal Energy – sum of all kinetic and potential energy in an object. It is very hard to determine an objects internal energy, but it is possible to determine the change in energy (  E). Change in internal energy,  E = E final – E initial –A positive  E means E final > E initial or the system gained energy from the surroundings (endothermic) –A negative  E means E final < E initial or the system lost energy to the surroundings (exothermic) Internal Energy

10. Chapter 510 Endothermic – a process which absorbs heat from the surroundings. Exothermic – a process which transfers heat to the surroundings. First Law of Thermodynamics Endothermic and Exothermic Processes

11. Chapter 511  E = q + w q = heatw = work Both heat energy and work can change a systems internal energy. First Law of Thermodynamics Relating  E to Heat and Work

12. Chapter 512 Relating  E to Heat and Work First Law of Thermodynamics  E = q + w

13. Chapter 513 “A process that is not path dependant.” Work (w) and heat (q) are not state functions. Energy change (  E) is a state function. First Law of Thermodynamics State Functions State function – a process that is determined by its initial and final conditions.

14. Chapter 514 State function – a process that is determined by its initial and final conditions. First Law of Thermodynamics State Functions

15. Chapter 515 State function – a process that is determined by its initial and final conditions. First Law of Thermodynamics State Functions

16. Chapter 516 Enthalpy (H) - Heat transferred between the system and surroundings carried out under constant pressure.  E = q + w Most reactions occur under constant pressure, so  E = q + (-P(  V))  V = 0 If volume is also constant,  V = 0  E = q p So, Energy change is due to heat transfer,  E =  H = q p Enthalpy

17. Chapter 517 Enthalpy Change (  H) – The heat evolved or absorbed in a reaction at constant pressure  H = H final - H initial = q P Enthalpy

18. Chapter 518 Enthalpy Change (  H) – The heat evolved or absorbed in a reaction at constant pressure  H = H final - H initial = q P Enthalpy

19. Chapter 519 Enthalpy Change (  H) – The heat evolved or absorbed in a reaction at constant pressure H and  H are state functions, depending only on the initial and final states. Enthalpy

20. Chapter 520 For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): Enthalpies of Reaction

21. Chapter 521 For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ Enthalpies of Reaction

22. Chapter 522 For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2CH 4 (g) + 4O 2 (g)  2CO 2 (g) + 4H 2 O(g)  H = -1604 kJ Enthalpies of Reaction

23. Chapter 523 For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: Enthalpies of Reaction

24. Chapter 524 For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: CO 2 (g) + 2H 2 O(g)  CH 4 (g) + 2O 2 (g)  H = +802 kJ Enthalpies of Reaction

25. Chapter 525 For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: CO 2 (g) + 2H 2 O(g)  CH 4 (g) + 2O 2 (g)  H = +802 kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ Enthalpies of Reaction

26. Chapter 526 For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: 3.Change in enthalpy depends on state: Enthalpies of Reaction

27. Chapter 527 For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: 3.Change in enthalpy depends on state: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g)  H = -802 kJ Enthalpies of Reaction

28. Chapter 528 For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: 3.Change in enthalpy depends on state: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g)  H = -802 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l)  H = -890 kJ Enthalpies of Reaction

29. Chapter 529 Enthalpies of Reaction

30. Chapter 530 a) Is this reaction endothermic or exothermic? Exothermic, this is indicated by the negative  H. Enthalpies of Reaction Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ

31. Chapter 531 b) Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Enthalpies of Reaction Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ

32. Chapter 532 b) Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Enthalpies of Reaction Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ

33. Chapter 533 b) Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Enthalpies of Reaction Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ

34. Chapter 534 b) Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Enthalpies of Reaction Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ

35. Chapter 535 b) Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Enthalpies of Reaction Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ

36. Chapter 536 Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Enthalpies of Reaction

37. Chapter 537 Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Enthalpies of Reaction

38. Chapter 538 Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Enthalpies of Reaction

39. Chapter 539 Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Enthalpies of Reaction

40. Chapter 540 Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Enthalpies of Reaction

41. Chapter 541 Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

42. Chapter 542 Problem: 5.33 2 MgO(s)  2 Mg(s) + O 2 (g)  H = 1204 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

43. Chapter 543 Problem: 5.33 2 MgO(s)  2 Mg(s) + O 2 (g)  H = 1204 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

44. Chapter 544 Problem: 5.33 2 MgO(s)  2 Mg(s) + O 2 (g)  H = 1204 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

45. Chapter 545 Problem: 5.33 2 MgO(s)  2 Mg(s) + O 2 (g)  H = 1204 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

46. Chapter 546 -Method of measuring heat flow and quantity of heat. Heat Capacity – The amount of heat required to raise an object’s temperature by 1 Kelvin. - As the heat capacity of a body increases, the amount of heat required to produce a given temperature change is increased. Calorimetry

47. Chapter 547 Molar Heat Capacity – The amount of heat required to raise the temperature of one mole of a substance by 1 Kelvin. Specific Heat (Specific Heat Capacity, c) – The amount of heat required to raise 1.0 gram of a substance by 1 Kelvin. Calorimetry

48. Chapter 548 Molar Heat Capacity – The amount of heat required to raise the temperature of one mole of a substance by 1 Kelvin. Specific Heat (Specific Heat Capacity, c) – The amount of heat required to raise 1.0 gram of a substance by 1 Kelvin. Calorimetry

49. Chapter 549 Molar Heat Capacity – The amount of heat required to raise the temperature of one mole of a substance by 1 Kelvin. Specific Heat (Specific Heat Capacity, c) – The amount of heat required to raise 1.0 gram of a substance by 1 Kelvin. Calorimetry

50. Chapter 550 Constant-Pressure Calorimetry Calorimetry

51. Chapter 551 - Atmospheric pressure is constant  H = q P q system = -q surroundings -The surroundings are composed of the water in the calorimeter and the calorimeter. q system = - (q water + q calorimeter ) -For most calculations, the q calorimeter can be ignored. q system = - q water c system m system  T system = - c water m water  T water Calorimetry Constant-Pressure Calorimetry

52. Chapter 552 Calorimetry Bomb Calorimetry (Constant-Volume Calorimetry)

53. Chapter 553 -Special calorimetry for combustion reactions -Substance of interest is placed in a “bomb” and filled to a high pressure of oxygen -The sealed bomb is ignited and the heat from the reaction is transferred to the water -This calculation must take into account the heat capacity of the calorimeter (this is grouped together with the heat capacity of water). q rxn = -C calorimeter (  T) Calorimetry Bomb Calorimetry (Constant-Volume Calorimetry)

54. Chapter 554 Problem 5.50 NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq)  T water = 18.4 o C – 23.0 o C = -4.6 o C m water = 60.0g c water = 4.184J/g o C m sample = 3.88g q sample = -q water q sample = -c water m water  T water q sample = -(4.184J/g o C)(60.0g+3.88g)(-4.6 o C) q sample = 1229J - Now calculate  H in kJ/mol Calorimetry

55. Chapter 555 Problem 5.50 NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq)  T water = 18.4 o C – 23.0 o C = -4.6 o C m water = 60.0g c water = 4.184J/g o C m sample = 3.88g Calorimetry q sample = 1229J moles NH 4 NO 3 = 3.88g/80.032g/mol = 0.04848mol  H = q sample /moles  H = 1229J/0.04848mol  H = 25.4 kJ/mol

56. Chapter 556 A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane? Calorimetry 2 C 8 H 18 + 25O 2  16 CO 2 + 18 H 2 O  T water = 28.78 o C – 21.36 o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g

57. Chapter 557 2 C 8 H 18 + 25O 2  16 CO 2 + 18 H 2 O  T water = 28.78 o C – 21.36 o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g q rxn = -C cal (  T water ) Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

58. Chapter 558 2 C 8 H 18 + 25O 2  16 CO 2 + 18 H 2 O  T water = 28.78 o C – 21.36 o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g q rxn = -C cal (  T water ) q rxn = -11.66kJ/ o C(7.42 o C) Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

59. Chapter 559 2 C 8 H 18 + 25O 2  16 CO 2 + 18 H 2 O  T water = 28.78 o C – 21.36 o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g q rxn = -C cal (  T water ) q rxn = -11.66kJ/ o C(7.42 o C) = -86.52kJ Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

60. Chapter 560  H combustion (in kJ/g)  H combustion = -86.52kJ/1.80g = Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

61. Chapter 561  H combustion (in kJ/g)  H combustion = -86.52kJ/1.80g = -48.1 kJ/g Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

62. Chapter 562  H combustion (in kJ/g)  H combustion = -86.52kJ/1.80g = -48.1 kJ/g  H combustion (in kJ/mol)  H combustion = -86.52kJ/0.01575mol = Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

63. Chapter 563  H combustion (in kJ/g)  H combustion = -86.52kJ/1.80g = -48.1 kJ/g  H combustion (in kJ/mol)  H combustion = -86.52kJ/0.01575mol = -5492 kJ/mol Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

64. Chapter 564 Hess’s law - If a reaction is carried out in a series of steps,  H for the overall reaction is the sum of  H’s for each individual step. Hess’s Law

65. Chapter 565 Hess’s law - If a reaction is carried out in a series of steps,  H for the overall reaction is the sum of  H’s for each individual step. For example: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2H 2 O(g)  2H 2 O(l)  H = -88 kJ Hess’s Law

66. Chapter 566 Hess’s law - If a reaction is carried out in a series of steps,  H for the overall reaction is the sum of  H’s for each individual step. For example: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2H 2 O(g)  2H 2 O(l)  H = -88 kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H = -890 kJ Hess’s Law

67. Chapter 567 Enthalpies of Formation (Heat of Formation) -There are many type of  H, depending on what you want to know  H vapor – enthalpy of vaporization (liquid  gas)  H fusion – enthalpy of fusion (solid  liquid)  H combustion – enthalpy of combustion (energy from burning a substance)

68. Chapter 568 Enthalpies of Formation (Heat of Formation) -A fundamental  H is the Standard Enthalpy of Formation ( )

69. Chapter 569 Enthalpies of Formation (Heat of Formation) -A fundamental  H is the Standard Enthalpy of Formation ( ) Standard Enthalpy of Formation ( ) – The enthalpy change that accompanies the formation of one mole of a substance from the most stable forms of its component elements at 298 Kelvin and 1 atmosphere pressure.

70. Chapter 570 Enthalpies of Formation (Heat of Formation) -A fundamental  H is the Standard Enthalpy of Formation ( ) Standard Enthalpy of Formation ( ) – The enthalpy change that accompanies the formation of one mole of a substance from the most stable forms of its component elements at 298 Kelvin and 1 atmosphere pressure. “The standard enthalpy of formation of the most stable form on any element is zero”

71. Chapter 571 Enthalpies of Formation

72. Chapter 572 Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction For a reaction:

73. Chapter 573 Enthalpies of Formation Problem 5.72 a) N 2 O 4 (g) + 4 H 2 (g)  N 2 (g) + 4 H 2 O(g) N 2 O 4 (g)9.66 kJ/mol H 2 (g)0 kJ/mol N 2 (g)0 kJ/mol H 2 O(g) -241.82 kJ/mol  H = [1mol(  H(N 2 )) + 4mol(  H(H 2 O))] – [1mol(  H(N 2 O 4 )) + 4mol(  H(H 2 ))]  H = [1mol(0kJ/mol) + 4mol(-241.82kJ/mol)] – [1mol(9.66kJ/mol) + 4mol(0kJ/mol)] = -976 kJ

74. Chapter 574 Enthalpies of Formation Problem 5.72 b) 2 KOH(s) + CO 2 (g)  K 2 CO 3 (s) + H 2 O(g) KOH(s) -424.7 kJ/mol CO 2 (g) -393.5 kJ/mol K 2 CO 3 (s) -1150.18 kJ/mol H 2 O(g) -241.82 kJ/mol  H = [1mol(  H(K 2 CO 3 )) + 1mol(  H(H 2 O))] – [2mol(  H(KOH)) + 1mol(  H(CO 2 ))]  H = [1mol(-1150.18kJ/mol) + 1mol(-241.82kJ/mol)] – [2mol(-424.7kJ/mol) + 1mol(-393.5kJ/mol)] = -149.1kJ

75. Chapter 575 5.20, 5.34, 5.36, 5.48, 5.54, 5.60, 5.70, 5.76 Practice Problems