1. Chapter 10 “Thermochemistry”

2. 2 Section 10.1 The Flow of Energy – Heat and Work u OBJECTIVES: Explain how energy, heat, and work are related.

3. 3 Section 10.1 The Flow of Energy – Heat and Work u OBJECTIVES: Identify the units used to measure heat transfer.

4. 4 Section 10.1 The Flow of Energy – Heat and Work u OBJECTIVES: Distinguish between heat capacity and specific heat capacity (also called simply specific heat).

5. 5 Energy Transformations u “Thermochemistry” - concerned with heat changes that occur during chemical reactions u Energy - capacity for doing work or supplying heat weightless, odorless, tasteless if within the chemical substances- called chemical potential energy

6. 6 Energy Transformations u Gasoline contains a significant amount of chemical potential energy u Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them. only changes can be detected! flows from warmer  cooler object

7. 7 Units for Measuring Heat Flow 1)A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 o C. Used except when referring to food a Calorie, (written with a capital C), always refers to the energy in food 1 Calorie = 1 kilocalorie = 1000 cal.

8. 8 Units for Measuring Heat Flow 2)The calorie is also related to the Joule, the SI unit of heat and energy named after James Prescott Joule 4.184 J = 1 cal u Heat Capacity - the amount of heat needed to increase the temperature of an object exactly 1 o C Depends on both the object’s mass and its chemical composition

9. 9 Heat Capacity and Specific Heat u Specific Heat Capacity (abbreviated “C”) - the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 o C often called simply “Specific Heat” Note Table 10.1, page 266 (next slide) u Water has a HUGE value, when it is compared to other chemicals

10. Table of Specific Heats Note the tremendous difference in Specific Heat. Water’s value is VERY HIGH.

11. 11 Heat Capacity and Specific Heat u For water, C = 4.18 J/(g o C) in Joules, and C = 1.00 cal/(g o C) in calories. u Thus, for water: it takes a long time to heat up, and it takes a long time to cool off! u Water is used as a coolant!

12. 12 Heat Capacity and Specific Heat u To calculate, use the formula: q = mass (in grams) x  T x C u heat is abbreviated as “q” u  T = change in temperature u C = Specific Heat Units are either: J/(g o C) or cal/(g o C)

13. - Page 510

14. u Try practice problem 10 & 11 on page 268 *Remember: q = mass (in grams) x  T x C Prob 10: C = q/mT = 42.8J/(deg Cx181g) = 0.236 J/(g o C) Prob 11: C = q/mT = 435J/(3.4 g x 64 o C) = 2.0 J/(g o C)

15. 15 Section 10.4 The Flow of Energy – Heat and Work u OBJECTIVES: Classify processes as either exothermic or endothermic.

16. Exothermic and Endothermic Processes u Essentially all chemical reactions and changes in physical state involve either: a)release of heat, or b)absorption of heat

17. Exothermic and Endothermic Processes u In studying heat changes, think of defining these two parts: the system - the part of the universe on which you focus your attention the surroundings - includes everything else in the universe

18. Exothermic and Endothermic Processes u Together, the system and it’s surroundings constitute the universe u Thermochemistry is concerned with the flow of heat from the system to it’s surroundings, and vice-versa.

19. Exothermic and Endothermic Processes u The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed. All the energy is accounted for as work, stored energy, or heat.

20. Exothermic and Endothermic Processes u Heat flowing into a system from it’s surroundings: defined as positive q has a positive value called endothermic –system gains heat (gets warmer/absorbs energy) as the surroundings cool down

21. Exothermic and Endothermic Processes u Heat flowing out of a system into it’s surroundings: defined as negative q has a negative value called exothermic –system loses heat (gets cooler/releases energy) as the surroundings heat up

22. 22 Exothermic and Endothermic u Every reaction has an energy change associated with it u ** Exothermic reactions release energy, usually in the form of heat. u Endothermic reactions absorb energy u Energy is stored in bonds between atoms

23. 23 Section 10.5 & 10.6 Measuring and Expressing Enthalpy Changes u OBJECTIVES: Describe how calorimeters are used to measure heat flow.

24. 24 Section 10.6 Measuring and Expressing Enthalpy Changes u OBJECTIVES: Construct thermochemical equations.

25. 25 Section 10.6 Measuring and Expressing Enthalpy Changes u OBJECTIVES: Solve for enthalpy changes in chemical reactions by using heats of reaction.

26. 10.5 Calorimetry u Calorimetry - the measurement of the heat into or out of a system for chemical and physical processes. Based on the fact that the heat released = the heat absorbed u The device used to measure the absorption or release of heat in chemical or physical processes is called a “Calorimeter”

27. Calorimetry u Foam cups are excellent heat insulators, and are commonly used as simple calorimeters under constant pressure. Fig. 10.5, page 270 u For systems at constant pressure, the “heat content” is the same as a property called Enthalpy (H) of the system (They are good because they are well-insulated.)

28. A foam cup calorimeter – here, two cups are nestled together for better insulation

29. Calorimetry u Changes in enthalpy =  H u q =  H These terms will be used interchangeably in this textbook u Thus, q =  H = m x C x  T 4  H is negative for an exothermic reaction 4  H is positive for an endothermic reaction

30. Calorimetry u Calorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed system Figure 10.6 page 270 Used by nutritionists to measure energy content of food

31. A Bomb Calorimeter A bomb calorimeter http://www.chm.davidson.edu/ronutt/che115/Bomb/Bomb.htm

32. - Page 513

33. 33 C + O 2 → CO 2 Energy ReactantsProducts  C + O 2 CO 2 395kJ given off + 395 kJ

34. 34 Exothermic u The products are lower in energy than the reactants u Thus, energy is released. u ΔH = -395 kJ The negative sign does not mean negative energy, but instead that energy is lost.

35. 35 CaCO 3 → CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 176 kJ absorbed CaCO 3 + 176 kJ → CaO + CO 2

36. 36 Endothermic u The products are higher in energy than the reactants u Thus, energy is absorbed. u ΔH = +176 kJ The positive sign means energy is absorbed

37. 37 Chemistry Happens in MOLES u An equation that includes energy is called a thermochemical equation  CH 4 + 2O 2  CO 2 + 2H 2 O + 802.2 kJ 1 mole of CH 4 releases 802.2 kJ of energy. When you make 802.2 kJ you also make 2 moles of water

38. 38 Thermochemical Equations u The heat of reaction is the heat change for the equation, exactly as written The physical state of reactants and products must also be given. Standard conditions (SC) for the reaction is 101.3 kPa (1 atm.) and 25 o C (different from STP)

39. 39 CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (l) + 802.2 kJ u If 10. 3 grams of CH 4 are burned completely, how much heat will be produced? 10. 3 g CH 4 16.05 g CH 4 1 mol CH 4 802.2 kJ = 514 kJ ΔH = -514 kJ, which means the heat is released for the reaction of 10.3 grams CH 4 Ratio from balanced equation 1 Start with known value Convert to moles Convert moles to desired unit

40. - Page 516

41. Summary, so far...

42. 42 Enthalpy u The heat content a substance has at a given temperature and pressure Can’t be measured directly because there is no set starting point u The reactants start with a heat content u The products end up with a heat content u So we can measure how much enthalpy changes

43. 43 Enthalpy u Symbol is H  Change in enthalpy is  H (delta H) u If heat is released, the heat content of the products is lower  H is negative (exothermic) u If heat is absorbed, the heat content of the products is higher  H is positive (endothermic)

44. 44 Energy ReactantsProducts  Change is down ΔH is <0 = Exothermic (heat is given off)

45. 45 Energy ReactantsProducts  Change is up ΔH is > 0 = Endothermic (heat is absorbed)

46. 46 Heat of Reaction u The heat that is released or absorbed in a chemical reaction  Equivalent to  H  C + O 2 (g)  CO 2 (g) + 393.5 kJ  C + O 2 (g)  CO 2 (g)  H = -393.5 kJ u In thermochemical equation, it is important to indicate the physical state a) H 2 (g) + 1/2O 2 (g)  H 2 O(g)  H = -241.8 kJ b) H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -285.8 kJ

47. 47 Heat of Combustion u The heat from the reaction that completely burns 1 mole of a substance:  C + O 2 (g)  CO 2 (g) + 393.5 kJ  C + O 2 (g)  CO 2 (g)  H = -393.5 kJ u Note Table 10.5, page 273 u Do practice problems 18, 19, 20 pg 274

48. 48 Section 10.7 Heat and Changes of State u OBJECTIVES: Describe the enthalpy change that occurs when a substance: a) melts, b) freezes, c) boils, d) condenses, or e) dissolves.

49. 49 Section 10.7 Heat and Changes of State u OBJECTIVES: Solve for the enthalpy change that occurs when a substance: a) melts, b) freezes, c) boils, d) condenses, or e) dissolves.

50. 50 Heat and Changes of State 1. Molar Heat of Fusion (  H fus. ) = the heat absorbed by one mole of a substance in melting from a solid to a liquid q = mol x  H fus. (no temperature change) Values given in Table 10.6, page 276 2. Molar Heat of Solidification (  H solid. ) = the heat lost when one mole of liquid solidifies (or freezes) to a solid q = mol x  H solid. (no temperature change)

51. 51 Heat in Changes of State u Note: You may also have the value of these equations as: q = mass x  H This is because some textbooks give the value of  H as kJ/gram, instead of kJ/mol

52. 52 Heat in Changes of State u Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies Thus,  H fus. = -  H solid. u Note Table 10.6, page 276 for the number values. Why is there no value listed for the molar heat of solidification?

53. Example 4 - Page 277 Do practice problem 22, pg 277

54. 54 Heats of Vaporization and Condensation u When liquids absorb heat at their boiling points, they become vapors. 3. Molar Heat of Vaporization (  H vap. ) = the amount of heat necessary to vaporize one mole of a given liquid. q = mol x  H vap. (no temperature change) u Table 10.6, page 276

55. 55 Heats of Vaporization and Condensation u Condensation is the opposite of vaporization. 4. Molar Heat of Condensation (  H cond. ) = amount of heat released when one mole of vapor condenses to a liquid q = mol x  H cond. (no temperature change) u  H vap. = -  H cond.

56. 56 Heats of Vaporization and Condensation u Lets look at Table 10.6, page 276… u The large values for water  H vap. and  H cond. is the reason hot vapors such as steam are very dangerous! You can receive a scalding burn from steam when the heat of condensation is released! H 2 0 (g)  H 2 0 (l)  H cond. = - 40.7kJ/mol

57. 57 The solid temperature is rising from -20 to 0 o C (use q = mol x ΔT x C) The solid is melting at 0 o C; no temperature change (use q = mol x ΔH fus. ) The liquid temperature is rising from 0 to 100 o C (use q = mol x ΔT x C) The liquid is boiling at 100 o C; no temperature change (use q = mol x ΔH vap. ) The gas temperature is rising from 100 to 120 o C (use q = mol x ΔT x C) The Heat Curve for Water, going from -20 to 120 o C, similar to the picture on page 523 120

58. Example 5- Page 278 Do practice problem 24, pg 278

59. 59 Heat of Solution u Heat changes can also occur when a solute dissolves in a solvent. 5. Molar Heat of Solution (  H soln. ) = heat change caused by dissolution of one mole of substance q = mol x  H soln. (no temperature change) u Sodium hydroxide provides a good example of an exothermic molar heat of solution (next slide)

60. 60 Heat of Solution NaOH (s)  Na 1+ (aq) + OH 1- (aq)  H soln. = - 445.1 kJ/mol u The heat is released as the ions separate (by dissolving) and interact with water, releasing 445.1 kJ of heat as  H soln. thus becoming so hot it steams! H 2 O (l)

62. - Example 6 Page 279

63. Practice Problem u Complete Practice Problem 25, pg 279 u Remember:  H soln. = 25.7 kJ/mol u Answer: 3.43 mol NH 4 NO 3

64. 10.7 Section Quiz. 1. The molar heat of condensation of a substance is the same, in magnitude, as its molar heat of A.formation. B.fusion. C.solidification. D.vaporization.

65. 2. The heat of condensation of ethanol (C 2 H 5 OH) is  43.5 kJ/mol. As C 2 H 5 OH condenses, the temperature of the surroundings A.stays the same. B.may increase or decrease. C.increases. D.decreases. 10.7 Section Quiz

66. 3. Calculate the amount of heat absorbed to liquefy 15.0 g of methanol (CH 3 OH) at its melting point. The molar heat of fusion for methanol is 3.16 kJ/mol A.1.48 kJ B.47.4 kJ C.1.52 10 3 kJ D.4.75 kJ

67. 4. How much heat (in kJ) is absorbed when 50 g of NH 4 NO 3 (s), 0.510 moles, are dissolved in water?  H soln = 25.7 kJ/mol A.12.85 kJ B.13.1 kJ C.25.7 kJ D.1285 kJ 10.7 Section Quiz

68. 68 Section 10.8 Calculating Heats of Reaction u OBJECTIVES: State Hess’s Law of Heat Summation, and describe how (or why) it is used in chemistry.

69. 69 Section 10.8 Calculating Heats of Reaction u OBJECTIVES: Solve for enthalpy changes by using Hess’s law or standard heats of formation.

70. 70 Hess’s Law (developed in 1840) u If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. Called Hess’s Law of Heat Summation Germain Henri Hess (1802-1850)

71. 71 How Does It Work? 1)If you turn an equation around, you change the sign: If H 2 (g) + 1/2 O 2 (g)  H 2 O(g)  H=-285.5 kJ then the reverse is: H 2 O(g)  H 2 (g) + 1/2 O 2 (g)  H =+285.5 kJ 2)If you multiply the equation by a number, you multiply the heat by that number: 2 H 2 O(g)  H 2 (g) + O 2 (g)  H =+571.0 kJ 3)Or, you can just leave the equation “as is”

72. 72 Hess’s Law - Procedure Options: 1. Use the equation as written 2. Reverse the equation (and change heat sign + to -, etc.) 3. Increase the coefficients in the equation (and increase heat by same amount) u Note samples from pages 281 and 282

73. 73 Standard Heats of Formation  The  H for a reaction that produces (or forms) 1 mol of a compound from its elements at standard conditions u Standard conditions: 25°C and 1 atm. u Symbol is: u The standard heat of formation of an element in it’s standard state is arbitrarily set at “ 0” u This includes the diatomic elements

74. 74 Standard Heats of Formation u Appendix Am Table A.6, page 787 has standard heats of formation u The heat of a reaction can be calculated by: subtracting the heats of formation of the reactants from the products HoHo = (Products) -(Reactants) Remember, from balanced equation: Products - Reactants

75. Standard Heats of Formation

76. - Page 531

77. 77 Another Example  CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) CH 4 (g) = - 74.86 kJ/molO 2 (g) = 0 kJ/molCO 2 (g) = - 393.5 kJ/molH 2 O(g) = - 241.8 kJ/mol  H= [-393.5 + 2(-241.8)] - [-74.86 +2 (0)]   H= - 802.24 kJ (endothermic or exothermic?) (Because it is an element )

78. –1. According to Hess’s law, it is possible to calculate an unknown heat of reaction by using A.heats of fusion for each of the compounds in the reaction. B.two other reactions with known heats of reaction. C.specific heat capacities for each compound in the reaction. D.density for each compound in the reaction. 10.8 & 10.9 Section Quiz.

79. –2. The heat of formation of Cl 2 (g) at 25°C is A.the same as that of H 2 O at 25°C. B.larger than that of Fe(s) at 25°C. C.undefined. D.zero.

80. 10.8 & 10.9 Section Quiz. 3.Calculate  H 0 for NH 3 (g) + HCl(g)  NH 4 Cl(s). Standard heats of formation: NH 3 (g) =  45.9 kJ/mol, HCl(g) =  92.3 kJ/mol, NH 4 Cl(s) =  314.4 kJ/mol A.176.2 kJ B.  360.8 kJ C.  176.2 kJ D.  268 kJ.

81. 81