1. Chapter Six 1 Hall © 2005 Prentice Hall © 2005 Thermochemistry Chapter ?

2. Chapter Six 2 Hall © 2005 Prentice Hall © 2005 Energy is the capacity to do work (to displace or move matter). Energy literally means “work within”; however, an object does not contain work. Potential energy is energy of position or composition. Kinetic energy is the energy of motion. E k = ½ mv 2 Energy has the units of joules (J or kg. m 2 /s 2 ) Energy

3. Chapter Six 3 Hall © 2005 Prentice Hall © 2005 Potential Energy and Kinetic Energy At what point in each bounce is the potential energy of the ball at a maximum?

4. Chapter Six 4 Hall © 2005 Prentice Hall © 2005 Thermochemistry is the study of energy changes that occur during chemical reactions. System: the part of the universe being studied. Surroundings: the rest of the universe. Thermochemistry: Basic Terms

5. Chapter Six 5 Hall © 2005 Prentice Hall © 2005 Open: energy and matter can be exchanged with the surroundings. Closed: energy can be exchanged with the surroundings, matter cannot. Isolated: neither energy nor matter can be exchanged with the surroundings. Types of Systems A closed system; energy (not matter) can be exchanged. After the lid of the jar is unscrewed, which kind of system is it?

6. Chapter Six 6 Hall © 2005 Prentice Hall © 2005 Internal energy (E) is the total energy contained within a system Part of E is kinetic energy (from molecular motion) –Translational motion, rotational motion, vibrational motion. –Collectively, these are sometimes called thermal energy Internal Energy (E) Part of E is potential energy –Intermolecular and intramolecular forces of attraction, locations of atoms and of bonds. –Collectively these are sometimes called chemical energy

7. Chapter Six 7 Hall © 2005 Prentice Hall © 2005 Technically speaking, heat is not “energy.” Heat is energy transfer between a system and its surroundings, caused by a temperature difference. More energetic molecules … … transfer energy to less energetic molecules. Heat (q) How do the root-mean-square speeds of the Ar atoms and the N 2 molecules compare at the point of thermal equilibrium? Thermal equilibrium occurs when the system and surroundings reach the same temperature and heat transfer stops.

8. Chapter Six 8 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic.

9. Chapter Six 9 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system into the surroundings, the process is exothermic.

10. Chapter Six 10 Hall © 2005 Prentice Hall © 2005 Like heat, work is an energy transfer between a system and its surroundings. Unlike heat, work is caused by a force moving through a distance (heat is caused by a temperature difference). A negative quantity of work signifies that the system loses energy. A positive quantity of work signifies that the system gains energy. There is no such thing as “negative energy” nor “positive energy”; the sign of work (or heat) signifies the direction of energy flow. Work (w)

11. Chapter Six 11 Hall © 2005 Prentice Hall © 2005 For now we will consider only pressure-volume work. work (w) = –P  V Pressure-Volume Work How would the magnitude of  V compare to the original gas volume if the two weights (initial and final) were identical?

12. Chapter Six 12 Hall © 2005 Prentice Hall © 2005 What is work? Work is a force acting over a distance. w= F x  d P = F/ area d = V/area w= (P x area) x  (V/area)= P  V Work can be calculated by multiplying pressure by the change in volume at constant pressure. units of liter - atm L-atm

13. Chapter Six 13 Hall © 2005 Prentice Hall © 2005 Work needs a sign If the volume of a gas increases, the system has done work on the surroundings. work is negative w = - P  V Expanding work is negative. Contracting, surroundings do work on the system w is positive. 1 L atm = 101.3 J

14. Chapter Six 14 Hall © 2005 Prentice Hall © 2005 Examples What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? If 2.36 J of heat are absorbed by the gas above. what is the change in energy?

15. Chapter Six 15 Hall © 2005 Prentice Hall © 2005 Same rules for heat and work Heat given off is negative. Heat absorbed is positive. Work done by system on surroundings is negative. Work done on system by surroundings is positive.

16. Chapter Six 16 Hall © 2005 Prentice Hall © 2005 The state of a system: its exact condition at a fixed instant. State is determined by the kinds and amounts of matter present, the structure of this matter at the molecular level, and the prevailing pressure and temperature. A state function is a property that has a unique value that depends only the present state of a system, and does not depend on how the state was reached (does not depend on the history of the system). Law of Conservation of Energy – in a physical or chemical change, energy can be exchanged between a system and its surroundings, but no energy can be created or destroyed. State Functions

17. Chapter Six 17 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its  E is the same. –But q and w are different in the two cases.

18. Chapter Six 18 Hall © 2005 Prentice Hall © 2005 “Energy cannot be created or destroyed.” Inference: the internal energy change of a system is simply the difference between its final and initial states:  E = E final – E initial Additional inference: if energy change occurs only as heat (q) and/or work (w), then:  E = q + w First Law of Thermodynamics

19. Chapter Six 19 Hall © 2005 Prentice Hall © 2005 Energy entering a system carries a positive sign: –heat absorbed by the system, or –work done on the system Energy leaving a system carries a negative sign –heat given off by the system –work done by the system First Law: Sign Convention

20. Chapter Six 20 Hall © 2005 Prentice Hall © 2005 Direction Every energy measurement has three parts. 1. A unit ( Joules of calories). 2. A number how many. 3. and a sign to tell direction. negative - exothermic positive- endothermic

21. Chapter Six 21 Hall © 2005 Prentice Hall © 2005 System Surroundings Energy  E <0

22. Chapter Six 22 Hall © 2005 Prentice Hall © 2005 System Surroundings Energy  E >0

23. Chapter Six 23 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is,  E = q + w.

24. Chapter Six 24 Hall © 2005 Prentice Hall © 2005 Example 6.1 A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat. What is the change in internal energy? Example 6.2: A Conceptual Example The internal energy of a fixed quantity of an ideal gas depends only on its temperature. If a sample of an ideal gas is allowed to expand against a constant pressure at a constant temperature, (a) what is  U for the gas? (b) Does the gas do work? (c) Is any heat exchanged with the surroundings?

25. Chapter Six 25 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc.  E, q, w, and Their Signs

26. Chapter Six 26 Hall © 2005 Prentice Hall © 2005 q rxn is the quantity of heat exchanged between a reaction system and its surroundings. An exothermic reaction gives off heat –In an isolated system, the temperature increases. –The system goes from higher to lower energy; q rxn is negative. An endothermic reaction absorbs heat –In an isolated system, the temperature decreases. –The system goes from lower to higher energy; q rxn is positive. Heats of Reaction (q rxn )

27. Chapter Six 27 Hall © 2005 Prentice Hall © 2005 Conceptualizing an Exothermic Reaction Surroundings are at 25 °C Hypothetical situation: all heat is instantly released to the surroundings. Heat = q rxn Typical situation: some heat is released to the surroundings, some heat is absorbed by the solution. In an isolated system, all heat is absorbed by the solution. Maximum temperature rise. 25 °C 32.2 °C 35.4 °C

28. Chapter Six 28 Hall © 2005 Prentice Hall © 2005 For a system where the reaction is carried out at constant volume,  V = 0 and  E = q V. Internal Energy Change at Constant Volume All the thermal energy produced by conversion from chemical energy is released as heat; no P-V work is done.

29. Chapter Six 29 Hall © 2005 Prentice Hall © 2005 lnternal Energy Change at Constant Pressure For a system where the reaction is carried out at constant pressure,  E = q P – P  V or  E + P  V = q P Most of the thermal energy is released as heat. Some work is done to expand the system against the surroundings (push back the atmosphere).

30. Chapter Six 30 Hall © 2005 Prentice Hall © 2005 We measure heat flow using calorimetry. A calorimeter is a device used to make this measurement. A “coffee cup” calorimeter may be used for measuring heat involving solutions. A “bomb” calorimeter is used to find heat of combustion; the “bomb” contains oxygen and a sample of the material to be burned. Calorimetry

31. Chapter Six 31 Hall © 2005 Prentice Hall © 2005 Calorimetry nMeasuring heat. nUse a calorimeter. nTwo kinds nConstant pressure calorimeter (called a coffee cup calorimeter) nheat capacity for a material, C is calculated C= heat absorbed/  T =  H/  T nspecific heat capacity = C/mass

32. Chapter Six 32 Hall © 2005 Prentice Hall © 2005 Heat evolved in a reaction is absorbed by the calorimeter and its contents. In a calorimeter we measure the temperature change of water or a solution to determine the heat absorbed or evolved by a reaction. The heat capacity (C) of a system is the quantity of heat required to change the temperature of the system by 1 ° C. C = q/  T (units are J/ ° C) Molar heat capacity is the heat capacity of one mole of a substance. The specific heat (s) is the heat capacity of one gram of a pure substance (or homogeneous mixture). s = C/m = q/(m  T) q = s m  T Calorimetry, Heat Capacity, Specific Heat

33. Chapter Six 33 Hall © 2005 Prentice Hall © 2005 Calorimetry nmolar heat capacity = C/moles heat = specific heat x m x  T heat = molar heat x moles x  T nMake the units work and you’ve done the problem right. A coffee cup calorimeter measures  H. nAn insulated cup, full of water. nThe specific heat of water is 1 cal/gºC Heat of reaction=  H = sh x mass x  T

34. Chapter Six 34 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1  C) is its heat capacity.

35. Chapter Six 35 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Heat Capacity and Specific Heat We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

36. Chapter Six 36 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Heat Capacity and Specific Heat Specific heat, then, is Specific heat = heat transferred mass  temperature change s = q m  Tm  T

37. Chapter Six 37 Hall © 2005 Prentice Hall © 2005 q = mass x specific heat x  T If  T is positive (temperature increases), q is positive and heat is gained by the system. If  T is negative (temperature decreases), q is negative and heat is lost by the system. The calorie, while not an SI unit, is still used to some extent. Water has a specific heat of 1 cal/(g o C). 4.184 J = 1 cal One food calorie (Cal or kcal) is actually equal to 1000 cal. More on Specific Heat

38. Chapter Six 38 Hall © 2005 Prentice Hall © 2005 Many metals have low specific heats. The specific heat of water is higher than that of almost any other substance.

39. Chapter Six 39 Hall © 2005 Prentice Hall © 2005 Heat Capacity: A Thought Experiment Place an empty iron pot weighing 5 lb on the burner of a stove. Place an iron pot weighing 1 lb and containing 4 lb water on a second identical burner (same total mass). Turn on both burners. Wait five minutes. Which pot handle can you grab with your bare hand? Iron has a lower specific heat than does water. It takes less heat to “warm up” iron than it does water.

40. Chapter Six 40 Hall © 2005 Prentice Hall © 2005 Example Calculate the heat capacity of an aluminum block that must absorb 629 J of heat from its surroundings in order for its temperature to rise from 22 ° C to 145 ° C.

41. Chapter Six 41 Hall © 2005 Prentice Hall © 2005 Example 6.7 How much heat, in joules and in kilojoules, does it take to raise the temperature of 225 g of water from 25.0 to 100.0 ° C? Example 6.8 What will be the final temperature if a 5.00-g silver ring at 37.0 ° C gives off 25.0 J of heat to its surroundings? Use the specific heat of silver listed in Table 6.1.

42. Chapter Six 42 Hall © 2005 Prentice Hall © 2005 Examples nThe specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K. nA 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

43. Chapter Six 43 Hall © 2005 Prentice Hall © 2005 Calorimetry nConstant volume calorimeter is called a bomb calorimeter. nMaterial is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. nThe heat capacity of the calorimeter is known and tested. Since  V = 0, P  V = 0,  E = q

44. Chapter Six 44 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

45. Chapter Six 45 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g-K), we can measure  H for the reaction with this equation: q = m  s   T

46. Chapter Six 46 Hall © 2005 Prentice Hall © 2005 Bomb Calorimeter nthermometer nstirrer nfull of water nignition wire nSteel bomb nsample

47. Chapter Six 47 Hall © 2005 Prentice Hall © 2005 Properties nintensive properties not related to the amount of substance. ndensity, specific heat, temperature. nExtensive property - does depend on the amount of stuff. nHeat capacity, mass, heat from a reaction.

48. Chapter Six 48 Hall © 2005 Prentice Hall © 2005 48 Enthalpy nSymbol is H Change in enthalpy is  H ndelta H nIf heat is released the heat content of the products is lower  H is negative (exothermic) nIf heat is absorbed the heat content of the products is higher  H is positive (endothermic)

49. Chapter Six 49 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Enthalpy Since  E = q + w and w = -P  V, we can substitute these into the enthalpy expression:  H =  E + P  V  H = (q+w) − w  H = q nSo, at constant pressure, the change in enthalpy is the heat gained or lost.

50. Chapter Six 50 Hall © 2005 Prentice Hall © 2005 Enthalpy is the sum of the internal energy and the pressure-volume product of a system: H = E + PV Enthalpy and Enthalpy Change Most reactions occur at constant pressure, so for most reactions, the heat evolved equals the enthalpy change. The evolved H 2 pushes back the atmosphere; work is done at constant pressure. For a process carried out at constant pressure, q P =  E + P  V soq P =  H Mg + 2 HCl  MgCl 2 + H 2

51. Chapter Six 51 Hall © 2005 Prentice Hall © 2005 Enthalpy is an extensive property. –It depends on how much of the substance is present. Since E, P, and V are all state functions, enthalpy H must be a state function also. Enthalpy changes have unique values.  H = q P Properties of Enthalpy Enthalpy change depends only on the initial and final states. In a chemical reaction we call the initial state the ____ and the final state the ____. Two logs on a fire give off twice as much heat as does one log.

52. Chapter Six 52 Hall © 2005 Prentice Hall © 2005 Values of  H are measured experimentally. Negative values indicate exothermic reactions. Positive values indicate endothermic reactions. Enthalpy Diagrams A decrease in enthalpy during the reaction;  H is negative. An increase in enthalpy during the reaction;  H is positive.

53. Chapter Six 53 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Enthalpy of Reaction The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants

54. Chapter Six 54 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Enthalpy of Reaction This quantity,  H, is called the enthalpy of reaction, or the heat of reaction.

55. Chapter Six 55 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. The Truth about Enthalpy 1.Enthalpy is an extensive property. 2.  H for a reaction in the forward direction is equal in size, but opposite in sign, to  H for the reverse reaction. 3.  H for a reaction depends on the state of the products and the state of the reactants.

56. Chapter Six 56 Hall © 2005 Prentice Hall © 2005  H changes sign when a process is reversed. Therefore, a cyclic process has the value  H = 0. Reversing a Reaction Same magnitude; different signs.

57. Chapter Six 57 Hall © 2005 Prentice Hall © 2005 Example 6.3 Given the equation (a) H 2 (g) + I 2 (s)  2 HI(g)  H = +52.96 kJ calculate  H for the reaction (b) HI(g)  ½ H 2 (g) + ½ I 2 (s). Example 6.4 The complete combustion of liquid octane, C 8 H 18, to produce gaseous carbon dioxide and liquid water at 25 °C and at a constant pressure gives off 47.9 kJ of heat per gram of octane. Write a chemical equation to represent this information.

58. Chapter Six 58 Hall © 2005 Prentice Hall © 2005 For problem-solving, heat evolved (exothermic reaction) can be thought of as a product. Heat absorbed (endothermic reaction) can be thought of as a reactant. We can generate conversion factors involving  H. For example, the reaction: ΔH in Stoichiometric Calculations H 2 (g) + Cl 2 (g)  2 HCl(g)  H = –184.6 kJ can be used to write: –184.6 kJ ———— 1 mol H 2 –184.6 kJ ———— 1 mol Cl 2 –184.6 kJ ———— 2 mol HCl

59. Chapter Six 59 Hall © 2005 Prentice Hall © 2005 Example 6.5 What is the enthalpy change associated with the formation of 5.67 mol HCl(g) in this reaction? H 2 (g) + Cl 2 (g)  2 HCl(g)  H = –184.6 kJ

60. Chapter Six 60 Hall © 2005 Prentice Hall © 2005 For a reaction carried out in a calorimeter, the heat evolved by a reaction is absorbed by the calorimeter and its contents. Measuring Enthalpy Changes for Chemical Reactions q rxn = – q calorimeter q calorimeter = mass x specific heat x  T By measuring the temperature change that occurs in a calorimeter, and using the specific heat and mass of the contents, the heat evolved (or absorbed) by a reaction can be determined and the enthalpy change calculated.

61. Chapter Six 61 Hall © 2005 Prentice Hall © 2005 Example 6.11 A 50.0-mL sample of 0.250 M HCl at 19.50 °C is added to 50.0 mL of 0.250 M NaOH, also at 19.50 °C, in a calorimeter. After mixing, the solution temperature rises to 21.21 °C. Calculate the heat of this reaction. Example 6.12 Express the result of Example 6.11 for molar amounts of the reactants and products. That is, determine the value of  H that should be written in the equation for the neutralization reaction: HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l)  H = ?

62. Chapter Six 62 Hall © 2005 Prentice Hall © 2005 Some reactions, such as combustion, cannot be carried out in a coffee-cup calorimeter. In a bomb calorimeter, a sample of known mass is placed in a heavy- walled “bomb,” which is then pressurized with oxygen. Since the reaction is carried out at constant volume, Bomb Calorimetry: Reactions at Constant Volume –q rxn = q calorimeter =  E … but in many cases the value of  E is a good approximation of  H.

63. Chapter Six 63 Hall © 2005 Prentice Hall © 2005 Some reactions cannot be carried out “as written.” Consider the reaction: C(graphite) + ½ O 2 (g)  CO(g). If we burned 1 mol C in ½ mol O 2, both CO and CO 2 would probably form. Some C might be left over. However … Hess’s Law of Constant Heat Summation

64. Chapter Six 64 Hall © 2005 Prentice Hall © 2005 … enthalpy change is a state function. The enthalpy change of a reaction is the same whether the reaction is carried out in one step or through a number of steps. Hess’s Law: If an equation can be expressed as the sum of two or more other equations, the enthalpy change for the desired equation is the sum of the enthalpy changes of the other equations. Hess’s Law of Constant Heat Summation

65. Chapter Six 65 Hall © 2005 Prentice Hall © 2005 Hess’s Law nEnthalpy is a state function. nIt is independent of the path. We can add equations to to come up with the desired final product, and add the  H nTwo rules If the reaction is reversed the sign of  H is changed If the reaction is multiplied, so is  H

66. Chapter Six 66 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Hess’s Law Hess’s law states that “[i]f a reaction is carried out in a series of steps,  H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

67. Chapter Six 67 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Hess’s Law Because  H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

68. Chapter Six 68 Hall © 2005 Prentice Hall © 2005 Example 6.14 Calculate the enthalpy change for reaction (a) given the data in equations (b), (c), and (d). (a) 2 C(graphite) + 2 H 2 (g)  C 2 H 4 (g)  H = ? (b) C(graphite) + O 2 (g)  CO 2 (g)  H = –393.5 kJ (c) C 2 H 4 (g) + 3 O 2  2 CO 2 (g) + 2 H 2 O(l)  H = –1410.9 kJ (d) H 2 (g) + ½ O 2  H 2 O(l)  H = –285.8 kJ

69. Chapter Six 69 Hall © 2005 Prentice Hall © 2005 Standard Enthalpies of Formation It would be convenient to be able to use the simple relationship ΔH = H products – H reactants to determine enthalpy changes. Although we don’t know absolute values of enthalpy, we don’t need them; we can use a relative scale. We define the standard state of a substance as the state of the pure substance at 1 atm pressure and the temperature of interest (usually 25 °C). The standard enthalpy change (ΔH°) for a reaction is the enthalpy change in which reactants and products are in their standard states. The standard enthalpy of formation (ΔH f °) for a reaction is the enthalpy change that occurs when 1 mol of a substance is formed from its component elements in their standard states.

70. Chapter Six 70 Hall © 2005 Prentice Hall © 2005 Standard Enthalpies of Formation nHess’s Law is much more useful if you know lots of reactions. nMade a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states. nStandard states are 1 atm, 1M and 25ºC nFor an element it is 0

71. Chapter Six 71 Hall © 2005 Prentice Hall © 2005 © 2009, Prentice-Hall, Inc. Standard Enthalpies of Formation Standard enthalpies of formation,  H f °, are measured under standard conditions (25 °C and 1.00 atm pressure).

72. Chapter Six 72 Hall © 2005 Prentice Hall © 2005 When we say “The standard enthalpy of formation of CH 3 OH(l) is –238.7 kJ”, we are saying that the reaction: C(graphite) + 2 H 2 (g) + ½ O 2 (g)  CH 3 OH(l) has a value of ΔH of –238.7 kJ. Standard Enthalpy of Formation We can treat ΔH f ° values as though they were absolute enthalpies, to determine enthalpy changes for reactions. Question: What is ΔH f ° for an element in its standard state [such as O 2 (g)]? Hint: since the reactants are the same as the products …

73. Chapter Six 73 Hall © 2005 Prentice Hall © 2005  H ° rxn =  p x  H f ° (products) –  r x  H f ° (reactants) The symbol  signifies the summation of several terms. The symbol signifies the stoichiometric coefficient used in front of a chemical symbol or formula. In other words … 1.Add all of the values for  H f ° of the products. 2.Add all of the values for  H f ° of the reactants. 3.Subtract #2 from #1 (This is usually much easier than using Hess’s Law!) Calculations Based on Standard Enthalpies of Formation

74. Chapter Six 74 Hall © 2005 Prentice Hall © 2005 Example 6.15 Synthesis gas is a mixture of carbon monoxide and hydrogen that is used to synthesize a variety of organic compounds. One reaction for producing synthesis gas is 3 CH 4 (g) + 2 H 2 O(l) + CO 2 (g)  4 CO(g) + 8 H 2 (g) ΔH° = ? Use standard enthalpies of formation from Table 6.2 to calculate the standard enthalpy change for this reaction. Example 6.16 The combustion of isopropyl alcohol, common rubbing alcohol, is represented by the equation 2 (CH 3 ) 2 CHOH(l) + 9 O 2 (g)  6 CO 2 (g) + 8 H 2 O(l) ΔH° = –4011 kJ Use this equation and data from Table 6.2 to establish the standard enthalpy of formation for isopropyl alcohol. Example 6.17: A Conceptual Example Without performing a calculation, determine which of these two substances should yield the greater quantity of heat per mole upon complete combustion: ethane, C 2 H 6 (g), or ethanol, CH 3 CH 2 OH(l).

75. Chapter Six 75 Hall © 2005 Prentice Hall © 2005 Ionic Reactions in Solution We can apply thermochemical concepts to reactions in ionic solution by arbitrarily assigning an enthalpy of formation of zero to H + (aq).

76. Chapter Six 76 Hall © 2005 Prentice Hall © 2005 Example 6.18 H + (aq) + OH – (aq)  H 2 O(l) ΔH° = –55.8 kJ Use the net ionic equation just given, together with ΔH f ° = 0 for H + (aq), to obtain ΔH f ° for OH – (aq).

77. Chapter Six 77 Hall © 2005 Prentice Hall © 2005 A reaction that occurs (by itself) when the reactants are brought together under the appropriate conditions is said to be spontaneous. A discussion of entropy is needed to fully understand the concept of spontaneity, and will be discussed in Chapter 17. A spontaneous reaction isn’t necessarily fast (rusting; diamond  graphite; etc. are slow). The difference between the tendency of a reaction to occur and the rate at which a reaction occurs will be discussed in Chapter 13. Looking Ahead