1. Chapter 6 Thermochemistry

2. Section 6.1 The Nature of Energy Thermochemistry Chapter 6

3. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 3  Capacity to do work or to produce heat.  Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed.  The total energy content of the universe is constant. Energy

4. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 4  Potential energy – energy due to position or composition.  Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity. Energy

5. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 5  In the initial position, ball A has a higher potential energy than ball B. Initial Position

6. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 6  After A has rolled down the hill, the potential energy lost by A (changed to kinetic energy) has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B (which is less than the P.E of A ) The temp. of the hill slightly increases. Final Position

7. Section 6.1 The Nature of Energy

8. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 8  System – part of the universe on which we wish to focus attention.  Surroundings – include everything else in the universe. Transfer of Energy

9. Section 6.1 The Nature of Energy System and Surrounding

10. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 10  Endothermic Reaction:  Heat flow is into a system.  Absorb energy from the surroundings.  Exothermic Reaction:  Energy flows out of the system.  Energy gained by the surroundings must be equal to the energy lost by the system. Chemical Energy

11. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 11 Is the freezing of water an endothermic or exothermic process? Explain. - remove energy in order to slow the molecules down to form a solid. CONCEPT CHECK!

12. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 12 Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. (system is underlined) a)Your hand gets cold when you touch ice. b)The ice gets warmer when you touch it. c)Water boils in a kettle being heated on a stove. d)Water vapor condenses on a cold pipe. e)Ice cream melts. Exo Endo Exo Endo CONCEPT CHECK!

13. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 13 For each of the following, define a system and its surroundings and give the direction of energy transfer. a)Methane is burning in a Bunsen burner in a laboratory. b)Water drops, sitting on your skin after swimming, evaporate. CONCEPT CHECK!

14. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 14 Hydrogen gas and oxygen gas react violently to form water. Explain.  Which is lower in energy: a mixture of hydrogen and oxygen gases, or water? CONCEPT CHECK!

15. Section 6.1 The Nature of Energy Thermodynamics  The study of energy and its interconversions is called thermodynamics.  Law of conservation of energy is often called the first law of thermodynamics. Copyright © Cengage Learning. All rights reserved 15

16. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 16 Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system.  To change the internal energy of a system: ΔE = q + w q represents heat w represents work Internal Energy

17. Section 6.1 The Nature of Energy Work vs Energy Flow Copyright © Cengage Learning. All rights reserved 17 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE

18. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 18  Thermodynamic quantities consist of two parts:  Number gives the magnitude of the change.  Sign indicates the direction of the flow. Internal Energy (magnitude and direction)

19. Section 6.1 The Nature of Energy Internal Energy  Sign reflects the system’s point of view.  Endothermic Process:  q is positive  Exothermic Process:  q is negative Copyright © Cengage Learning. All rights reserved 19

20. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 20  Sign reflects the system’s point of view.  System does work on surroundings:  w is negative  Surroundings do work on the system:  w is positive Internal Energy

21. Section 6.1 The Nature of Energy 21  Work = P × A × Δh = PΔV  P is pressure.  A is area.  Δh is the piston moving a distance.  ΔV is the change in volume. Work

22. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 22  For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs: w = –PΔV  To convert between L·atm and Joules, use 1 L·atm = 101.3 J. Work

23. Section 6.1 The Nature of Energy Copyright © Cengage Learning. All rights reserved 23 Which of the following performs more work? a)A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. They perform the same amount of work. EXERCISE!

24. Section 6.2 Enthalpy and Calorimetry Change in Enthalpy  State function  ΔH = q at constant pressure  ΔH = H products – H reactants  Energy is a state function; work and heat are not  State Function – property that does not depend in any way on the system’s past or future (only depends on present state). Copyright © Cengage Learning. All rights reserved 24

25. Section 6.2 Enthalpy and Calorimetry Consider the combustion of propane: C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure. –252 kJ Copyright © Cengage Learning. All rights reserved 25 EXERCISE!

26. Section 6.2 Enthalpy and Calorimetry Calorimetry  Science of measuring heat  Specific heat capacity:  The energy required to raise the temperature of one gram of a substance by one degree Celsius.  Molar heat capacity:  The energy required to raise the temperature of one mole of substance by one degree Celsius. Copyright © Cengage Learning. All rights reserved 26

27. Section 6.2 Enthalpy and Calorimetry HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C. Which has the larger heat capacity?

28. Section 6.2 Enthalpy and Calorimetry specific heat(s) and heat capacity (C)  The specific heat(s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius  The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.  C = m x s  Heat (q) absorbed or released  q = m x s x  t  q = C x  t   t = t final - t initial

29. Section 6.2 Enthalpy and Calorimetry specific heat(s) and heat capacity (C)

30. Section 6.2 Enthalpy and Calorimetry Calorimetry  If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.  An endothermic reaction cools the solution. Copyright © Cengage Learning. All rights reserved 30

31. Section 6.2 Enthalpy and Calorimetry A Coffee–Cup Calorimeter Made of Two Styrofoam Cups Copyright © Cengage Learning. All rights reserved 31 RReview Expt.11RReview Expt.11

32. Section 6.2 Enthalpy and Calorimetry 32 Constant-Volume Calorimetry No heat enters or leaves! q q rxn = - (q water + q bomb ) rxn = - (q water + q bomb ) q water = m x s x  t q bomb = C bomb x  t Rea q water = m x s x  t ction at Constant V  H ~ rxn Reaction at Constant V  H ~ q rxn

33. Section 6.2 Enthalpy and Calorimetry Calorimetry  Energy released (heat) = s × m × ΔT s = specific heat capacity (J/°C·g) m = mass of solution (g) ΔT = change in temperature (°C) Copyright © Cengage Learning. All rights reserved 33

34. Section 6.2 Enthalpy and CalorimetryENTHALPYENTHALPY Most chemical reactions occur at constant P, so and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = H final - H initial and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = H final - H initial Heat transferred at constant P = q p q p = ∆H where H = enthalpy Heat transferred at constant P = q p q p = ∆H where H = enthalpy

35. Section 6.2 Enthalpy and Calorimetry A 466-g sample of water is heated from 8.50°C to 74.60°C. Calculate the amount of heat absorbed (in kilojoules) by the water.

36. Section 6.2 Enthalpy and Calorimetry Strategy We know the quantity of water and the specific heat of water. With this information and the temperature rise, we can calculate the amount of heat absorbed (q). Solution Using Equation (6.12), we write Check The units g and °C cancel, and we are left with the desired unit kJ. Because heat is absorbed by the water from the surroundings, it has a positive sign.

37. Section 6.2 Enthalpy and Calorimetry 17.How many kJ of heat must be removed from 1000 g of water (specific heat = 4.184 J /g o C) to lower the temperature from 18.0°C to 12.0°C? A. 2.5 x 10 –2 kJB. 1.4 kJC. 4.2 kJ D. 25 kJ q= ms∆t = 1000g x 4.184 j/ g o C x 6 o C =25104J = 25 kJ q= q= ms∆t = 1000g x 4.184 j/ g o C x 6 o C =25104J = 25 kJ ms∆t = 1000g x 4.184 j/ g o C x 6 o C =25104J = 25 kJ

38. Section 6.2 Enthalpy and Calorimetry A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Copyright © Cengage Learning. All rights reserved 38 CONCEPT CHECK!

39. Section 6.2 Enthalpy and Calorimetry A 100.0 g sample of water at 90.°C is added to a 500.0 g sample of water at 10.°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 23°C Copyright © Cengage Learning. All rights reserved 39 CONCEPT CHECK!

40. Section 6.2 Enthalpy and Calorimetry You have a Styrofoam cup with 50.0 g of water at 10.°C. You add a 50.0 g iron ball at 90. °C to the water. (s H2O = 4.18 J/°C·g and s Fe = 0.45 J/°C·g) The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 18°C Copyright © Cengage Learning. All rights reserved 40 CONCEPT CHECK!

41. Section 6.2 Enthalpy and Calorimetry Problem  A 1.000 g sample of rocket fuel hydrazine, N 2 H 4, is burned in a bomb calorimeter containing 1200 g of water. The temperature rises from 24.62 to 28.16 ⁰C. Taking C for the bomb to be 840 J/ ⁰C, calculate (a)Q reaction for the combustion of the one-gram sample. (b)Q reaction for the combustion of the one mole of hydrazine in the bomb calorimeter.

42. Section 6.2 Enthalpy and Calorimetry QAAaction = - (qwater + q calorimeter) q Q reaction = - (q water + q calorimeter ) q water = 1200 g x 4.184 J/g ⁰C x (28.16 – 24.62) ⁰C = 17773.6 J q calorimeter = 840J/ ⁰C x (28.16-24.62) = 2973.6 J (a) Q reaction for 1 g N 2 H 4 = 17773.6 J + 2973.6 J = 20747.2 J = -20.7 kJ (b) Q reaction for 1 mole of N 2 H 4 = 20.7 kJ /gx 32 g/mole = -662.4 kJ water – 1200 g x 4.184 J/g ⁰ C x (28.16 – 24.62) ⁰ C = 17773.6 J qcalorimeter = 840J/ ⁰ C x (28.16-24.62) = 2973.6 J Qreaction for 1 g N2H4 = 17773.6 J + 2973.6 J = 20747.2 J = -20.7 kJ Qreaction for 1 mole of N2H4 = 20.7 kJ /gx 32 g/mole = -662.4 kJ

43. Section 6.3 Hess’s Law  In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. Copyright © Cengage Learning. All rights reserved 43

44. Section 6.3 Hess’s Law  This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH 2 and ΔH 3. N 2 (g) + O 2 (g) → 2NO(g) ΔH 2 = 180 kJ 2NO(g) + O 2 (g) → 2NO 2 (g) ΔH 3 = – 112 kJ N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH 2 + ΔH 3 = 68 kJ ΔH 1 = ΔH 2 + ΔH 3 = 68 kJ N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH 1 = 68 kJ Copyright © Cengage Learning. All rights reserved 44

45. Section 6.3 Hess’s Law The Principle of Hess’s Law Copyright © Cengage Learning. All rights reserved 45

46. Section 6.3 Hess’s Law Copyright © Cengage Learning. All rights reserved 46 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE

47. Section 6.3 Hess’s Law Characteristics of Enthalpy Changes  If a reaction is reversed, the sign of ΔH is also reversed.  The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer. Copyright © Cengage Learning. All rights reserved 47

48. Section 6.3 Hess’s Law Example  Consider the following data:  Calculate ΔH for the reaction Copyright © Cengage Learning. All rights reserved 48

49. Section 6.3 Hess’s Law Problem-Solving Strategy  Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal.  Reverse any reactions as needed to give the required reactants and products.  Multiply reactions to give the correct numbers of reactants and products. Copyright © Cengage Learning. All rights reserved 49

50. Section 6.3 Hess’s Law Example  Reverse the two reactions:  Desired reaction: Copyright © Cengage Learning. All rights reserved 50

51. Section 6.3 Hess’s Law Example  Multiply reactions to give the correct numbers of reactants and products: 4( ) 4( ) 3( ) 3( )  Desired reaction:

52. Section 6.3 Hess’s Law Example  Final reactions:  Desired reaction: ΔH = +1268 kJ Copyright © Cengage Learning. All rights reserved 52

53. Section 6.3 Hess’s Law Could you use these data to obtain the enthalpy change for the following reaction? Hess’s Law you use these data to obtain the ealpy change for the following reaction? Fo For example, suppose you are given the following data: r example, suppose you are given the following data:

54. Section 6.3 Hess’s Law  If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third. Hess’s Law

55. Section 6.4 Standard Enthalpies of Formation Standard Enthalpy of Formation (ΔH f °)  Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. Copyright © Cengage Learning. All rights reserved 55

56. Section 6.4 Standard Enthalpies of Formation

57. Section 6.4 Standard Enthalpies of Formation

58. Section 6.4 Standard Enthalpies of Formation A Schematic Diagram of the Energy Changes for the Reaction CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) ∆H ◦ f of the compounds involved are: CH4: –75 kJ, CO2: -394 kJ, H2O: –286 kJ ∆H ◦ rxn = ∆H ◦ f products - ∆H ◦ f reactants = {[-394 + 2 x (-286)] –(-75)} = –891 kJ Answer: –891 kJ Copyright © Cengage Learning. All rights reserved 58

59. Section 6.5 Present Sources of Energy  Fossil Fuels  Petroleum, Natural Gas, and Coal  Wood  Hydro  Nuclear Copyright © Cengage Learning. All rights reserved 59

60. Section 6.5 Present Sources of Energy Energy Sources Used in the United States Copyright © Cengage Learning. All rights reserved 60

61. Section 6.5 Present Sources of Energy The Earth’s Atmosphere  Transparent to visible light from the sun.  Visible light strikes the Earth, and part of it is changed to infrared radiation.  Infrared radiation from Earth’s surface is strongly absorbed by CO 2, H 2 O, and other molecules present in smaller amounts in atmosphere.  Atmosphere traps some of the energy and keeps the Earth warmer than it would otherwise be. Copyright © Cengage Learning. All rights reserved 61

62. Section 6.5 Present Sources of Energy The Earth’s Atmosphere Copyright © Cengage Learning. All rights reserved 62

63. Section 6.6 New Energy Sources  Coal Conversion  Hydrogen as a Fuel  Other Energy Alternatives  Oil shale  Ethanol  Methanol  Seed oil Copyright © Cengage Learning. All rights reserved 63