Chapter 6 Thermochemistry

Transcription

1 Chapter 6 Thermochemistry

2 Contents and Concepts Understanding Heats of Reaction The first part of the chapter lays the groundwork for understanding what we mean by heats of reaction. 1. Energy and Its Units 2. Heat of Reaction 3. Enthalpy and Enthalpy Changes 4. Thermochemical Equations 5. Applying Stoichiometry to Heats of Reaction 6. Measuring Heats of Reaction Copyright Cengage Learning. All rights reserved. 6 2

3 Using Heats of Reaction Now that we understand the basic properties of heats of reaction and how to measure them, we can explore how to use them. 7. Hess s Law 8. Standard Enthalpies of Formation 9. Fuels Foods, Commercial Fuels, and Rocket Fuels Copyright Cengage Learning. All rights reserved. 6 3

4 Learning Objectives Understanding Heats of Reaction 1. Energy and Its Units a. Define energy, kinetic energy, and internal energy. b. Define the SI unit of energy (joule) as well as the common unit of energy (calorie). c. Calculate the kinetic energy of a moving object. d. State the law of conservation of energy. Copyright Cengage Learning. All rights reserved. 6 4

5 2. Heat of Reaction a. Define a thermodynamic system and its surroundings. b. Define heat and heat of reaction. c. Distinguish between an exothermic process and an endothermic process. 3. Enthalpy and Enthalpy Changes a. Define enthalpy and enthalpy of reaction. b. Explain how the terms enthalpy of reaction and heat of reaction are related. c. Explain how enthalpy and internal energy are related. Copyright Cengage Learning. All rights reserved. 6 5

6 4. Thermochemical Equations a. Define a thermochemical equation. b. Write a thermochemical equation given pertinent information. c. Learn the two rules for manipulating (reversing and multiplying) thermochemical equations. d. Manipulate a thermochemical equation using these rules. 5. Applying Stoichiometry to Heats of Reaction a. Calculate the heat absorbed or evolved from a reaction given its enthalpy of reaction and the mass of a reactant or product. Copyright Cengage Learning. All rights reserved. 6 6

7 6. Measuring Heats of Reaction a. Define heat capacity and specific heat. b. Relate the heat absorbed or evolved to the specific heat, mass, and temperature change. c. Perform calculations using the relationship between heat and specific heat. d. Define a calorimeter. e. Calculate the enthalpy of reaction from calorimetric data (temperature change and heat capacity). Copyright Cengage Learning. All rights reserved. 6 7

8 Using Heats of Reaction 7. Hess s Law a. State Hess s law of heat summation. b. Apply Hess s law to obtain the enthalpy change for one reaction from the enthalpy changes of a number of other reactions. Copyright Cengage Learning. All rights reserved. 6 8

9 8. Standard Enthalpies of Formation a. Define standard state and reference form. b. Define standard enthalpy of formation. c. Calculate the heat of a phase transition using standard enthalpies of formation for the different phases. d. Calculate the heat (enthalpy) of reaction from the standard enthalpies of formation of the substances in the reaction. Copyright Cengage Learning. All rights reserved. 6 9

10 9. Fuels Foods, Commercial Fuels, and Rocket Fuels a. Define a fuel. b. Describe the three needs of the body that are fulfilled by foods. c. Give the approximate average values quoted (per gram) for the heat values (heats of combustion) for fats and for carbohydrates. d. List the three major fossil fuels. e. Describe the processes of coal gasification and coal liquefaction. f. Describe some fuel-oxidizer systems used in rockets. Copyright Cengage Learning. All rights reserved. 6 10

11 Thermodynamics The science of the relationship between heat and other forms of energy. Thermochemistry An area of thermodynamics that concerns the study of the heat absorbed or evolved by a chemical reaction Copyright Cengage Learning. All rights reserved. 6 11

12 Energy The potential or capacity to move matter. One form of energy can be converted to another form of energy: electromagnetic, mechanical, electrical, or chemical. Next, we ll study kinetic energy, potential energy, and internal energy. Copyright Cengage Learning. All rights reserved. 6 12

13 Kinetic Energy, E K The energy associated with an object by virtue of its motion. E K = 1 2 mv m = mass (kg) v = velocity (m/s) 2 Copyright Cengage Learning. All rights reserved. 6 13

14 The SI unit of energy is the joule, J, pronounced jewel. 2 kg m J = 2 s The calorie is a non-si unit of energy commonly used by chemists. It was originally defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius. The exact definition is given by the equation: 1 cal = J (exact) Copyright Cengage Learning. All rights reserved. 6 14

15 ? A person weighing 75.0 kg (165 lbs) runs a course at 1.78 m/s (4.00 mph). What is the person s kinetic energy? m = 75.0 kg V = 1.78 m/s E E K = 1 (75.0 kg) K = kg m = s (3 significant figures) E K = ½ mv 2 m s J 2 Copyright Cengage Learning. All rights reserved. 6 15

16 Potential Energy, E P The energy an object has by virtue of its position in a field of force, such as gravitaitonal, electric or magnetic field. Gravitational potential energy is given by the equation E P = mgh m = mass (kg) g = gravitational constant (9.80 m/s 2 ) h = height (m) Copyright Cengage Learning. All rights reserved. 6 16

17 Internal Energy, U The sum of the kinetic and potential energies of the particles making up a substance. Total Energy E tot = E K + E P + U Copyright Cengage Learning. All rights reserved. 6 17

18 Law of Conservation of Energy Energy may be converted from one form to another, but the total quantity of energy remains constant. Copyright Cengage Learning. All rights reserved. 6 18

19 Thermodynamic System The substance under study in which a change occurs is called the thermodynamic system (or just system). Thermodynamic Surroundings Everything else in the vicinity is called the thermodynamic surroundings (or just the surroundings). Copyright Cengage Learning. All rights reserved. 6 19

20 Heat, q The energy that flows into or out of a system because of a difference in temperature between the thermodynamic system and its surroundings. Heat flows spontaneously from a region of higher temperature to a region of lower temperature. q is defined as positive if heat is absorbed by the system (heat is added to the system) q is defined as negative if heat is evolved by a system (heat is subtracted from the system) Copyright Cengage Learning. All rights reserved. 6 20

21 Heat of Reaction The value of q required to return a system to the given temperature at the completion of the reaction (at a given temperature) Copyright Cengage Learning. All rights reserved. 6 21

22 Endothermic Process A chemical reaction or process in which heat is absorbed by the system (q is positive). The reaction vessel will feel cool. Exothermic Process A chemical reaction or process in which heat is evolved by the system (q is negative). The reaction vessel will feel warm. Copyright Cengage Learning. All rights reserved. 6 22

23 In an endothermic reaction: The reaction vessel cools. Heat is absorbed. Energy is added to the system. q is positive. In an exothermic reaction: The reaction vessel warms. Heat is evolved. Energy is subtracted from the system. q is negative. Copyright Cengage Learning. All rights reserved. 6 23

24 Enthalpy, H An extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction. Extensive Property A property that depends on the amount of substance. Mass and volume are extensive properties. Copyright Cengage Learning. All rights reserved. 6 24

25 A state function is a property of a system that depends only on its present state, which is determined by variables such as temperature and pressure, and is independent of any previous history of the system. Copyright Cengage Learning. All rights reserved. 6 25

26 The altitude of a campsite is a state function. It is independent of the path taken to reach it. Copyright Cengage Learning. All rights reserved. 6 26

27 Enthalpy of Reaction The change in enthalpy for a reaction at a given temperature and pressure: H = H(products) H(reactants) Note: means change in. Enthalpy change is equal to the heat of reaction at constant pressure: H = q P Copyright Cengage Learning. All rights reserved. 6 27

28 The diagram illustrates the enthalpy change for the reaction 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) The reactants are at the top. The products are at the bottom. The products have less enthalpy than the reactants, so enthalpy is evolved as heat. The signs of both q and H are negative. Copyright Cengage Learning. All rights reserved. 6 28

29 Enthalpy and Internal Energy The precise definition of enthalpy, H, is H = U + PV Many reactions take place at constant pressure, so the change in enthalpy can be given by H = U + P V Rearranging: U = H P V The term ( P V) is the energy needed to change volume against the atmospheric pressure, P. It is called pressure-volume work. Copyright Cengage Learning. All rights reserved. 6 29

30 For the reaction 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) P V The H 2 gas had to do work to raise the piston. For the reaction as written at 1 atm, -P V = -2.5 kj. In addition, kj of heat are evolved. Copyright Cengage Learning. All rights reserved. 6 30

31 Thermochemical Equation The thermochemical equation is the chemical equation for a reaction (including phase labels) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation. For the reaction of sodium metal with water, the thermochemical equation is: 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g); H = kj Copyright Cengage Learning. All rights reserved. 6 31

32 ? Sulfur, S 8, burns in air to produce sulfur dioxide. The reaction evolves 9.31 kj of heat per gram of sulfur at constant pressure. Write the thermochemical equation for this reaction. Copyright Cengage Learning. All rights reserved. 6 32

33 We first write the balanced chemical equation: S 8 (s) + 8O 2 (g) 8SO 2 (g) Next, we convert the heat per gram to heat per mole. H 9.31kJ g S = 1g S 1mol S H 8 = Note: The negative sign indicates that heat is evolved; the reaction is exothermic. Now we can write the thermochemical equation: S 8 (s) + 8O 2 (g) 8SO 2 (g); H = kj 3 kj 8 8 Copyright Cengage Learning. All rights reserved. 6 33

34 Manipulating a Thermochemical Equation When the equation is multiplied by a factor, the value of H must be multiplied by the same factor. When a chemical equation is reversed, the sign of H is reversed. Copyright Cengage Learning. All rights reserved. 6 34

35 Copyright Cengage Learning. All rights reserved. 6 35

36 CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g); H = 206 kj a. CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) This reaction is identical to the given reaction. It is endothermic. H = 206 kj Copyright Cengage Learning. All rights reserved. 6 36

37 CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g); H = 206 kj b. 2 CH 4 (g) + 2H 2 O(g) 2CO(g) + 6H 2 (g) This reaction is double the given reaction. It is endothermic. H = 412 kj Copyright Cengage Learning. All rights reserved. 6 37

38 CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g); H = 206 kj c. CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) This reaction is the reverse of the given reaction. It is exothermic. H = -206 kj Copyright Cengage Learning. All rights reserved. 6 38

39 CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g); H = 206 kj d. 2CO(g) + 6H 2 (g) 2CH 4 (g) + 2H 2 O(g) This reaction is reverse and double the given reaction. It is exothermic. H = -412 kj Equations c and d are exothermic. Equation d is the most exothermic reaction. Copyright Cengage Learning. All rights reserved. 6 39

40 ? When sulfur burns in air, the following reaction occurs: S 8 (s) + 8O 2 (g) 8SO 2 (g); H = 2.39 x 10 3 kj Write the thermochemical equation for the dissociation of one mole of sulfur dioxide into its elements. Copyright Cengage Learning. All rights reserved. 6 40

41 S 8 (s) + 8O 2 (g) 8SO 2 (g); H = kj We want SO 2 as a reactant, so we reverse the given reaction, changing the sign of H: 8SO 2 (g) S 8 (g) + 8O 2 (g) ; H = kj We want only one mole SO 2, so now we divide every coefficient and H by 8: SO 2 (g) 1 / 8 S 8 (g) + O 2 (g) ; H = +299 kj Copyright Cengage Learning. All rights reserved. 6 41

42 Applying Stoichiometry to Heats of Reaction Copyright Cengage Learning. All rights reserved. 6 42

43 ? You burn 15.0 g sulfur in air. How much heat evolves from this amount of sulfur? The thermochemical equation is S 8 (s) + 8O 2 (g) 8SO 2 (g); H = x 10 3 kj Copyright Cengage Learning. All rights reserved. 6 43

44 S 8 (s) + 8O 2 (g) 8SO 2 (g); H = x 10 3 kj Molar mass of S 8 = g q = 15.0 g S 8 1mol S g S x 10 1mol S 8 3 kj q = kj Copyright Cengage Learning. All rights reserved. 6 44

45 The daily energy requirement for a 20- year-old man weighing 67 kg is 1.3 x 10 4 kj. For a 20-year-old woman weighing 58 kg, the daily requirement is 8.8 x 10 3 kj. If all this energy were to be provided by the combustion of glucose, C 6 H 12 O 6, how many grams of glucose would have to be consumed by the man and the woman per day? C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l); H = x 10 3 kj? Copyright Cengage Learning. All rights reserved. 6 45

46 C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l); H = x 10 3 kj For a 20-year-old man weighing 67 kg: 4 1mol glucose g glucose mglucose = 1.3x10 kj x10 kj 1mol glucose = 830 g glucose required (2 significant figures) For a 20-year-old woman weighing 58 kg: 3 1mol glucose g glucose mglucose = 8.8x10 kj x10 kj 1mol glucose = 560 g glucose required (2 significant figures) Copyright Cengage Learning. All rights reserved. 6 46

47 Measuring Heats of Reaction We will first look at the heat needed to raise the temperature of a substance because this is the basis of our measurements of heats of reaction. Copyright Cengage Learning. All rights reserved. 6 47

48 Heat Capacity, C, of a Sample of Substance The quantity of heat needed to raise the temperature of the sample of substance by one degree Celsius (or one Kelvin). Molar Heat Capacity The heat capacity for one mole of substance. Specific Heat Capacity, s (or specific heat) The quantity of heat needed to raise the temperature of one gram of substance by one degree Celsius (or one Kelvin) at constant pressure. Copyright Cengage Learning. All rights reserved. 6 48

49 The heat required can be found by using the following equations. Using heat capacity: q = C t Using specific heat capacity: q = s x m x t Copyright Cengage Learning. All rights reserved. 6 49

50 A calorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change. Two examples are shown below. Copyright Cengage Learning. All rights reserved. 6 50

51 ? A piece of zinc weighing 35.8 g was heated from C to C. How much heat was required? The specific heat of zinc is J/(g C). m = 35.8 g s = J/(g C) t = C C = 8.00 C q = J 35.8 g 8.00 C g C ( ) q = 111 J (3 significant figures) q = m s t Copyright Cengage Learning. All rights reserved. 6 51

52 ? Nitromethane, CH 3 NO 2, an organic solvent burns in oxygen according to the following reaction: CH 3 NO 2 (g) + 3 / 4 O 2 (g) CO 2 (g) + 3 / 2 H 2 O(l) + 1 / 2 N 2 (g) You place g of nitromethane in a calorimeter with oxygen and ignite it. The temperature of the calorimeter increases from C to C. The heat capacity of the calorimeter was determined to be kj/ C. Write the thermochemical equation for the reaction. Copyright Cengage Learning. All rights reserved. 6 52

53 We first find the heat evolved for the g of nitromethane, CH 3 NO 2. q rxn = Now, covert that to the heat evolved per mole by using the molar mass of nitromethane, g. q rxn = kj C q rxn ( C C) = kj kj g CH 3 = NO C 2 cal t H = 709 kj g CH3NO 1mol CH NO Copyright Cengage Learning. All rights reserved. 6 53

54 We can now write the thermochemical equation: CH 3 NO 2 (l) + ¾O 2 (g) CO 2 (g) + 3 / 2 H 2 O(l) + ½N 2 (g); H = 709 kj Copyright Cengage Learning. All rights reserved. 6 54

55 Hess s Law of Heat Summation For a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps. Copyright Cengage Learning. All rights reserved. 6 55

56 Suppose we want H for the reaction 2C(graphite) + O 2 (g) 2CO(g) It is difficult to measure directly. However, two other reactions are known: C(graphite) + O 2 (g) CO 2 (g); H = kj 2CO 2 (g) 2CO(g) + O 2 (g); H = kj In order for these to add to give the reaction we want, we must multiply the first reaction by 2. Note that we also multiply H by 2. Copyright Cengage Learning. All rights reserved. 6 56

57 2C(graphite) + O 2 (g) 2CO(g) 2C(graphite) + 2O 2 (g) 2CO 2 (g); H = kj 2CO 2 (g) 2CO(g) + O 2 (g); H = kj 2 C(graphite) + O 2 (g) 2 CO(g); H = kj Now we can add the reactions and the H values. Copyright Cengage Learning. All rights reserved. 6 57

58 H sub = H fus + H vap Copyright Cengage Learning. All rights reserved. 6 58

59 ? What is the enthalpy of reaction, H, for the reaction of calcium metal with water? Ca(s) + 2H 2 O(l) Ca 2+ (aq) + 2OH - (aq) + H 2 (g) This reaction occurs very slowly, so it is impractical to measure H directly. However, the following facts are known: H + (aq) + OH - (aq) H 2 O(l); H = 55.9 kj Ca(s) + 2H + (aq) Ca 2+ (aq) + H 2 (g); H = kj Copyright Cengage Learning. All rights reserved. 6 59

60 Ca(s) + 2H 2 O(l) Ca 2+ (aq) + 2OH - (aq) + H 2 (g) First, identify each reactant and product: H + (aq) + OH - (aq) H 2 O(l); H = 55.9 kj Ca(s) + 2H + (aq) Ca 2+ (aq) + H 2 (g); H = kj Each substance must be on the proper side. Ca(s), Ca 2+ (aq), and H 2 (g) are fine. H 2 O(l) should be a reactant. OH - (aq) should be a product. Reversing the first reaction and changing the sign of its H accomplishes this. Copyright Cengage Learning. All rights reserved. 6 60

61 Ca(s) + 2H 2 O(l) Ca 2+ (aq) + 2OH - (aq) + H 2 (g) H 2 O(l) H + (aq) + OH - (aq); H = kj Ca(s) + 2H + (aq) Ca 2+ (aq) + H 2 (g); H = kj The coefficients must match those in the reaction we want. The coefficient on H 2 O and OH - should be 2. We multiply the first reaction and its H by 2 to accomplish this. Copyright Cengage Learning. All rights reserved. 6 61

62 Ca(s) + 2H 2 O(l) Ca 2+ (aq) + 2OH - (aq) + H 2 (g) 2H 2 O(l) 2H + (aq) + 2OH - (aq); H = kj Ca(s) + 2H + (aq) Ca 2+ (aq) + H 2 (g); H = kj We can now add the equations and their H s. Note that 2H + (aq) appears as both a reactant and a product. Copyright Cengage Learning. All rights reserved. 6 62

63 Ca(s) + 2H 2 O(l) Ca 2+ (aq) + 2OH - (aq) + H 2 (g) 2H 2 O(l) 2H + (aq) + 2OH - (aq); H = kj Ca(s) + 2H + (aq) Ca 2+ (aq) + H 2 (g); H = kj Ca(s) + 2H 2 O(l) Ca 2+ (aq) + 2OH - (aq) + H 2 (g); H = kj Copyright Cengage Learning. All rights reserved. 6 63

64 Standard Enthalpies of Formation The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atm pressure and the specified temperature (usually 25 C). These standard conditions are indicated with a degree sign ( ). When reactants in their standard states yield products in their standard states, the enthalpy of reaction is called the standard enthalpy of reaction, H. ( H is read delta H zero. ) Copyright Cengage Learning. All rights reserved. 6 64

65 Elements can exist in more than one physical state, and some elements exist in more than one distinct form in the same physical state. For example, carbon can exist as graphite or as diamond; oxygen can exist as O 2 or as O 3 (ozone). These different forms of an element in the same physical state are called allotropes. The reference form is the most stable form of the element (both physical state and allotrope). Copyright Cengage Learning. All rights reserved. 6 65

66 The standard enthalpy of formation, H f, is the enthalpy change for the formation of one mole of the substance from its elements in their reference forms and in their standard states. H f for an element in its reference and standard state is zero. For example, the standard enthalpy of formation for liquid water is the enthalpy change for the reaction H 2 (g) + 1 / 2 O 2 (g) H 2 O(l) H f = kj Other H f values are given in Table 6.2 and Appendix C. Copyright Cengage Learning. All rights reserved. 6 66

67 Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, H. CH 4 (g) + 4Cl 2 (g) CCl 4 (l) + 4HCl(g); H =? Table 6.2 shows the H f values: C(graphite) + 2Cl 2 (g) CCl 4 (l); H f = kj 1 / 2 H 2 (g) + 1 / 2 Cl 2 (g) HCl(g); H f = kj CH 4 (g) C(graphite) + 2H 2 (g); H f = kj We first identify each reactant and product from the reaction we want. Copyright Cengage Learning. All rights reserved. 6 67

68 Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, H. CH 4 (g) + 4Cl 2 (g) CCl 4 (l) + 4HCl(g); H =? Table 6.2 shows the H f values: C(graphite) + 2Cl 2 (g) CCl 4 (l); H f = kj 1 / 2 H 2 (g) + 1 / 2 Cl 2 (g) HCl(g); H f = kj CH 4 (g) C(graphite) + 2H 2 (g); H f = kj Each needs to be on the correct side of the arrow and is. Next, we ll check coefficients. Copyright Cengage Learning. All rights reserved. 6 68

69 Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, H. CH 4 (g) + 4Cl 2 (g) CCl 4 (l) + 4HCl(g); H =? Table 6.2 shows the H f values: C(graphite) + 2Cl 2 (g) CCl 4 (l); H f = kj 1 / 2 H 2 (g) + 1 / 2 Cl 2 (g) HCl(g); H f = kj CH 4 (g) C(graphite) + 2H 2 (g); H f = kj Cl 2 and HCl need a coefficient of 4. Multiplying the second equation and its H by 4 does this. Copyright Cengage Learning. All rights reserved. 6 69

70 Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, H. CH 4 (g) + 4Cl 2 (g) CCl 4 (l) + 4HCl(g); H =? Table 6.2 shows the H f values: C(graphite) + 2Cl 2 (g) CCl 4 (l); H f = kj 2H 2 (g) + 2Cl 2 (g) 4HCl(g); H f = kj CH 4 (g) C(graphite) + 2H 2 (g); H f = kj Now, we can add the equations. Copyright Cengage Learning. All rights reserved. 6 70

71 Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, H. CH 4 (g) + 4Cl 2 (g) CCl 4 (l) + 4HCl(g); H =? Table 6.2 shows the H f values: C(graphite) + 2Cl 2 (g) CCl 4 (l); H f = kj 2H 2 (g) + 2Cl 2 (g) 4HCl(g); H f = kj CH 4 (g) C(graphite) + 2H 2 (g); H f = kj CH 4 (g) + 4Cl 2 (g) CCl 4 (g) + 4HCl(g); H = kj Copyright Cengage Learning. All rights reserved. 6 71

72 ? What is the heat of vaporization of methanol, CH 3 OH, at 25 C and 1 atm? Use standard enthalpies of formation (Appendix C). Copyright Cengage Learning. All rights reserved. 6 72

73 We want H for the reaction: CH 3 OH(l) CH 3 OH(g) For For H reaction = products n H kj liquid methanol : H f = mol kj gaseous methanol : H f = mol f reactants n H f H vap = 1mol kj mol 1mol kj mol H vap = kj Copyright Cengage Learning. All rights reserved. 6 73

74 Methyl alcohol, CH? 3 OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO, which can coagulate protein. Calculate H o for the following reaction: 2CH 3 OH(aq) + O 2 (g) 2HCHO(aq) + 2H 2 O(l) o Standard enthalpies of formation, H f : CH 3 OH(aq): kj/mol HCHO(aq): kj/mol H 2 O(l): kj/mol Copyright Cengage Learning. All rights reserved. 6 74

75 We want H for the reaction: 2CH 3 OH(aq) + O 2 (aq) 2HCHO(aq) + 2H 2 O(l) H o reacton H H o reaction = reaction 2 mol 2 mol = = products n H kj mol kj mol f + reactants + 1mol 0 n H kj mol f 2 mol kj mol [ ( kj) + ( kj) ] [ kj] H o reaction = ( kj) [ kj] H o reaction = kj Copyright Cengage Learning. All rights reserved. 6 75

76 Foods fuels three needs of the body: They supply substances for the growth and repair of tissue. They supply substances for the synthesis of compounds used in the regulation processes. They supply energy. Foods do this by a combustion process. Copyright Cengage Learning. All rights reserved. 6 76

77 For glucose, a carbohydrate: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l); H f = 2803 kj For glycerol trimyristate, a fat: C 45 H 86 O 6 (s) / 2 O 2 (g) 45CO 2 (g) + 43H 2 O(l); H f = 27,820 kj The average value for carbohydrates is 4.0 kcal/g and for fats is 9.0 kcal/g. Copyright Cengage Learning. All rights reserved. 6 77

78 Fossil fuels originated millions of years ago when aquatic plants and animals were buried and compressed by layers of sediment at the bottoms of swamps and seas. Over time this organic matter was converted by bacterial decay and pressure to petroleum (oil), gas, and coal. Copyright Cengage Learning. All rights reserved. 6 78

79 Coal, which accounts for 22.9% of total U.S. energy consumption, varies in terms of the amount of carbon it contains and so varies in terms of the amount of energy it produces in combustion. Anthracite (hard coal) was laid down as long as 250 million years ago and can contain as much as 80% carbon. Bituminous coal, a younger variety, contains between 45% and 65% carbon. Copyright Cengage Learning. All rights reserved. 6 79

80 Natural gas, which accounts for 22.7% of total U.S. energy consumption, is convenient because it is fluid and can be easily transported. Purified natural gas is primarily methane, CH 4, plus small amounts of ethane, C 2 H 6 ; propane, C 3 H 8 ; and butane, C 4 H 10. Petroleum is a mixture of compounds. Gasoline, which is obtained from petroleum, is a mixture of hydrocarbons (compounds of carbon and hydrogen). Copyright Cengage Learning. All rights reserved. 6 80

81 The main issue with natural gas and petroleum is their relatively short supply. It has been estimated that petroleum deposits will be 80% depleted by Natural gas deposits may be depleted even sooner. Coal deposits, however, are expected to last for several more centuries. This has led to the development of commercial methods for converting coal to the more easily handled liquid and gaseous fuels. Copyright Cengage Learning. All rights reserved. 6 81

82 Coal gasification is one way. Steam is passed over hot coal: C(s) + H 2 O(g) CO(g) + H 2 (g) The mixture containing carbon monoxide can be converted by various methods into useful products. The mixture of carbon monoxide and hydrogen can be converted by various methods into useful products such as methane, CH 4. CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) Copyright Cengage Learning. All rights reserved. 6 82

83 Rockets are self-contained missiles propelled by the ejection of gases from an orifice. Usually these are hot gases expelled from the rocket from the reaction of a fuel with an oxidizer. One factor in determining the appropriate fuel/oxidizer combination is its mass of the mixture. Copyright Cengage Learning. All rights reserved. 6 83