1. Chapter 17 - Thermochemistry -is the study of energy relationships in chemical and physical reactions

2. u almost nothing happens in chemistry unless there is an energy advantage; all reactions are an attempt to arrive at a lower system energy.

3. a) 1st law of Thermodynamics: Law of conservation of energy. u Energy is never lost or gained, it only changes form. This law states that energy can change form but is never really lost in any closed system. b) 2nd law of Thermodynamics: u Heat Energy always travels spontaneously from a warmer body (body with higher temperature) to a colder body. -droping a hot rock in water -frostbite

4. 4 u OBJECTIVES: Explain the relationship between energy and heat. Distinguish between heat capacity and specific heat.

5. u temperature is simply an indirect measure of the average kinetic energy of the particles in a sample of matter. The scale of measurement is arbitrary: u i)Celsius scale sets the zero value at the freezing point of water at 101.325 kPa and 100 at the boiling point of water at the same pressure. u ii)Kelvin scale sets the zero value at absolute zero. u The degrees are the same size as celsius degrees.

6. 6 Energy and Heat 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

7. 7 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

8. Exothermic and Endothermic Processes u Essentially all chemical reactions, and changes in physical state, involve either: release of heat, (exothermic) or absorption of heat (endothermic)

9. 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

10. 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.

11. 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.

12. u heat flowing into a system from it’s surroundings: defined as positive q has a positive value called endothermic –system gains heat as the surroundings cool down

13. u heat flowing out of a system into it’s surroundings: defined as negative q has a negative value called exothermic –system loses heat as the surroundings heat up

14. 14 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

15. 15 Heat Capacity and Specific Heat u 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.

16. 16 Heat Capacity and Specific Heat u 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

17. 17 Heat Capacity and Specific Heat u Specific Heat Capacity - 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” u mass; measured in g or kg. The greater the mass, the more heat required to change the temperature.

18. 18 u For water, C = 4.18 J/(g o C), 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!

19. 19 Heat Capacity and Specific Heat u To calculate, use the formula: u q = m C  T u heat abbreviated as “q” u m = mass u  T = change in temperature ( Tf - Ti) u C = Specific Heat u Units are either J/(g o C) or kJ/(kg o C)

20. Table of Specific Heats

21. 21 Ex: Determine the heat required to raise the temperature of 100g of water from 298.0 K to 373.0 K. Q = m c ΔT Q = 100g (4.184 J/g K)( 373.0 K – 298.0 K) Q = 418.4 J/K (75 K) Q = 31350 J Q = 31.4 kJ

22. 22

23. Calorimetry u Calorimetry - the accurate and precise measurement of heat change for chemical and physical processes. u The device used to measure the absorption or release of heat in chemical or physical processes is called a Calorimeter

24. Calorimetry u Foam cups are excellent heat insulators, and are commonly used as simple calorimeters

25. A Cheap Calorimeter

26. u For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system

27. 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 u  H is negative for an exothermic reaction u  H is positive for an endothermic reaction

28. Calorimetry u Calorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed system

30. 30 C + O 2  CO 2 Energy ReactantsProducts  C + O 2 C O 2 395kJ + 395 kJ

31. 31 In terms of bonds C O O C O O Breaking this bond will require energy. C O O O O C Making these bonds gives you energy. In this case making the bonds gives you more energy than breaking them.

32. 32 Exothermic u The products are lower in energy than the reactants u Releases energy

33. 33 CaCO 3  CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 176 kJ CaCO 3 + 176 kJ  CaO + CO 2

34. 34 Endothermic u The products are higher in energy than the reactants u Absorbs energy

35. 35 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 u 1 mole of CH 4 releases 802.2 kJ of energy. u When you make 802.2 kJ you also make 2 moles of water

36. u What is the molar enthalpy of CO2 (g) in the reaction for the burning of butane below? 2 C 4 H 10 +13 O 2  8 CO 2 +10 H 2 O ∆H=-5315 kJ Answer: Molar enthalpy is the enthalpy change in equation divided by the balance of CO2 u Molar enthalpy, H substance = 5315 kJ ÷ 8 mol u = 664 kJ / mol.

37. 37 Thermochemical Equations u A 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 for the reaction is 101.3 kPa (1 atm.) and 25 o C

38. 38 CH 4 + 2 O 2  CO 2 + 2 H 2 O + 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

39. 39 CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u How many liters of O 2 at STP would be required to produce 23 kJ of heat? u How many grams of water would be produced with 506 kJ of heat?

40. u How much heat will be released if 65 grams of butane is burned in a lighter according the equation: 2 C 4 H 10 +13 O 2  8 CO 2 +10 H 2 O ∆H=-5315 kJ = 2976.4 kJ = 3.0 MJ

41. u Calculate the heat released when 120 grams of Iron (III) oxide is formed by the following equation 2 Fe 2 O 3 (s) → 4 Fe(s)+3 O 2 (g) ∆H=1625 kJ = 610.5 kJ = 610 kJ

42. u Q = n ∆H (substance) Where n = # of moles

43. u What mass of carbon dioxide must form to create 1200 kJ of heat when the following reaction occurs? C 6 H 12 O 6 (s)+6O 2 (g)→6CO 2 (g)+6H 2 O(l) ∆H=- 2808kJ Answer: 110 grams

44. u 3) What mass of oxygen is needed to completely react and release 550 kJ of heat in the following reaction? u 4Fe (s)+3O 2 (g) → 2 Fe 2 O 3 (s) ∆H=- 1625 kJ Answer: 32 grams

45. Summary, so far...

46. 46 Enthalpy u The heat content a substance has at a given temperature and pressure u 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

47. 47 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)

48. 48 Energy ReactantsProducts  Change is down  H is <0

49. 49 Energy ReactantsProducts  Change is up  H is > 0

50. 50 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  H 2 (g) + 1/2O 2 (g)  H 2 O(g)  H = -241.8 kJ  H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -285.8 kJ

51. 51 Heat of Combustion u The heat from the reaction that completely burns 1 mole of a substance

52. 52 u OBJECTIVES: Classify, by type, the heat changes that occur during melting, freezing, boiling, and condensing.

53. 53 u OBJECTIVES: Calculate heat changes that occur during melting, freezing, boiling, and condensing.

54. 54 Heats of Fusion and Solidification u Molar Heat of Fusion (  H fus ) - the heat absorbed by one mole of a substance in melting from a solid to a liquid u Molar Heat of Solidification (  H solid ) - heat lost when one mole of liquid solidifies

55. 55 Heats of Fusion and Solidification u Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies Thus,  H fus = -  H solid

56. 56 Heats of Vaporization and Condensation u When liquids absorb heat at their boiling points, they become vapors. u Molar Heat of Vaporization (  H vap ) - the amount of heat necessary to vaporize one mole of a given liquid.

57. 57 Heats of Vaporization and Condensation u Condensation is the opposite of vaporization. u Molar Heat of Condensation (  H cond ) - amount of heat released when one mole of vapor condenses u  H vap = -  H cond

58. 58 Heats of Vaporization and Condensation The large values for  H vap and  H cond are the reason hot vapors such as steam is very dangerous You can receive a scalding burn from steam when the heat of condensation is released!

59. 59 Heats of Vaporization and Condensation u H 2 0 (g)  H 2 0 (l)  H cond = - 40.7kJ/mol

60. 60 Heat of Solution u Heat changes can also occur when a solute dissolves in a solvent. u Molar Heat of Solution (  H soln ) - heat change caused by dissolution of one mole of substance u Sodium hydroxide provides a good example of an exothermic molar heat of solution:

61. 61 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 and interact with water, releasing 445.1 kJ of heat as  H soln thus becoming so hot it steams H 2 O (l)

62. 62 u OBJECTIVES: Apply Hess’s law of heat summation to find heat changes for chemical and physical processes.

63. 63 u OBJECTIVES: Calculate heat changes using standard heats of formation.

64. 64 Hess’s Law 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

65. 65 Why Does It Work? u 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, H 2 O(g)  H 2 (g) + 1/2 O 2 (g)  H =+285.5 kJ u also, u 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

66. 66 u You make the products, so you need their heats of formation u You “unmake” the products so you have to subtract their heats. u How do you get good at this?

67. 67 Standard Heats of Formation  The  H for a reaction that produces 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 = 0 u This includes the diatomics

68. 68 What good are they? u Table 11.6, page 316 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)

69. 69 Examples  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)] - [-74.68 +2 (0)]  H= - 802.4 kJ

70. 70 Examples