7.9: Exercises

Questions

7.2: Energy Basics

1. A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not?

2. Explain the difference between heat capacity and specific heat of a substance.

3. Calculate the heat capacity, in joules and in calories per degree, of the following:

1. 28.4 g of water
2. 1.00 oz of lead

4. Calculate the heat capacity, in joules and in calories per degree, of the following:

1. 45.8 g of nitrogen gas
2. 1.00 pound of aluminum metal

5. How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C?

6. How much heat, in joules and in calories, is required to heat a 28.4-g (1-oz) ice cube from −23.0 °C to −1.0 °C?

7. How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added?

8. If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?

9. A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from 23.2 °C to 89.6 °C.

1. What is the specific heat of the substance?
2. If it is one of the substances found in Table (Table 5.1), what is its likely identity?

10. A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.

1. What is the specific heat of the substance?
2. If it is one of the substances found in Table (Table 5.1), what is its likely identity?

11. An aluminum kettle weighs 1.05 kg.

1. What is the heat capacity of the kettle?
2. How much heat is required to increase the temperature of this kettle from 23.0 °C to 99.0 °C?
3. How much heat is required to heat this kettle from 23.0 °C to 99.0 °C if it contains 1.25 L of water (density of 0.997 g/mL and a specific heat of 4.184 J/g °C)?

12. Most people find waterbeds uncomfortable unless the water temperature is maintained at about 85 °F. Unless it is heated, a waterbed that contains 892 L of water cools from 85 °F to 72 °F in 24 hours. Estimate the amount of electrical energy required over 24 hours, in kWh, to keep the bed from cooling. Note that 1 kilowatt-hour (kWh) = 3.6 × 10⁶ J, and assume that the density of water is 1.0 g/mL (independent of temperature). What other assumptions did you make? How did they affect your calculated result (i.e., were they likely to yield “positive” or “negative” errors)?

7.3: Calorimetry

13. A 500-mL bottle of water at room temperature and a 2-L bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the 500-mL bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the 500-mL bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers.

14. Would the amount of heat measured for the reaction in Example (Example 5.5) be greater, lesser, or remain the same if we used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer.

15. Would the amount of heat absorbed by the dissolution in Example (Example 5.6) appear greater, lesser, or remain the same if the experimenter used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer.

16. Would the amount of heat absorbed by the dissolution in Example (Example 5.6) appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer.

17. How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.

18. How much will the temperature of a cup (180 g) of coffee at 95 °C be reduced when a 45 g silver spoon (specific heat 0.24 J/g °C) at 25 °C is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water.

19. A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of coffee at 85 °C and the temperature of the two become equal.

1. What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water.
2. The first time a student solved this problem she got an answer of 88 °C. Explain why this is clearly an incorrect answer.

20. The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiator it has a temperature of 175 °F. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of 4.184 J/g °C.

21. A 70.0-g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter like that shown in Figure (Figure 5.12). The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water? What is the specific heat of the metal?

22. If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5 °C in a calorimeter like that in Figure (Figure 5.12), what is the resulting temperature of the water?

23. A 0.500-g sample of KCl is added to 50.0 g of water in a calorimeter (Figure 5.12). If the temperature decreases by 1.05 °C, what is the approximate amount of heat involved in the dissolution of the KCl, assuming the heat capacity of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic?

24. Dissolving 3.0 g of CaCl₂(s) in 150.0 g of water in a calorimeter (Figure 5.12) at 22.4 °C causes the temperature to rise to 25.8 °C. What is the approximate amount of heat involved in the dissolution, assuming the heat capacity of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic?

25. When 50.0 g of 0.200 M NaCl(aq) at 24.1 °C is added to 100.0 g of 0.100 M AgNO₃(aq) at 24.1 °C in a calorimeter, the temperature increases to 25.2 °C as AgCl(s) forms. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat in joules produced.

26. The addition of 3.15 g of Ba(OH)₂•8H₂O to a solution of 1.52 g of NH₄SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:

\[Ba(OH)_2 \cdot 8H_2O_{(s)} + 2NH_4SCN_{(aq)} \rightarrow Ba(SCN)_{2(aq)} + 2NH_{3(aq)} + 10H_2O_{(l)}\]

27. The reaction of 50 mL of acid and 50 mL of base described in Example (Example 5.5) increased the temperature of the solution by 6.9 degrees. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 °C? Explain your answer.

28. If the 3.21 g of NH₄NO₃ in Example (Example 5.6) were dissolved in 100.0 g of water under the same conditions, how much would the temperature change? Explain your answer.

29. When 1.0 g of fructose, C₆H₁₂O₆(s), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is q for this combustion?

30. When a 0.740-g sample of trinitrotoluene (TNT), C₇H₅N₂O₆, is burned in a bomb calorimeter, the temperature increases from 23.4 °C to 26.9 °C. The heat capacity of the calorimeter is 534 J/°C, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample?

31. One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00 g of coal is burned in a bomb calorimeter, the temperature increases by 1.48 °C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat produced by combustion of a ton of coal (2.000 × 10³ pounds).

32. The amount of fat recommended for someone with a daily diet of 2000 Calories is 65 g. What percent of the calories in this diet would be supplied by this amount of fat if the average number of Calories for fat is 9.1 Calories/g?

33. A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories (16 kcal). What is the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g?

34. What is the maximum mass of carbohydrate in a 6-oz serving of diet soda that contains less than 1 Calorie per can if the average number of Calories for carbohydrates is 4.1 Calories/g?

35. A pint of premium ice cream can contain 1100 Calories. What mass of fat, in grams and pounds, must be produced in the body to store an extra 1.1 × 10³ Calories if the average number of Calories for fat is 9.1 Calories/g?

36. A serving of a breakfast cereal contains 3 g of protein, 18 g of carbohydrates, and 6 g of fat. What is the Calorie content of a serving of this cereal if the average number of Calories for fat is 9.1 Calories/g, for carbohydrates is 4.1 Calories/g, and for protein is 4.1 Calories/g?

37. Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs $4.23, or a liter of isooctane (density, 0.6919 g/mL) that costs $0.45? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories.

7.4: Enthalpy

38. Explain how the heat measured in Example 5.5 differs from the enthalpy change for the exothermic reaction described by the following equation:

\[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l)\]

39. Using the data in the check your learning section of Example 5.5, calculate ΔH in kJ/mol of AgNO₃(aq) for the reaction:

\[\ce{NaCl}(aq)+\ce{AgNO3}(aq)⟶\ce{AgCl}(s)+\ce{NaNO3}(aq)\]

40. Calculate the enthalpy of solution (ΔH for the dissolution) per mole of NH₄NO₃ under the conditions described in Example 5.6.

41. Calculate ΔH for the reaction described by the equation.

\(\ce{Ba(OH)2⋅8H2O}(s)+\ce{2NH4SCN}(aq)⟶\ce{Ba(SCN)2}(aq)+\ce{2NH3}(aq)+\ce{10H2O}(l)\)

42. Calculate the enthalpy of solution (ΔH for the dissolution) per mole of CaCl₂.

43. Although the gas used in an oxyacetylene torch is essentially pure acetylene, the heat produced by combustion of one mole of acetylene in such a torch is likely not equal to the enthalpy of combustion of acetylene listed in Table (Table 5.2). Considering the conditions for which the tabulated data are reported, suggest an explanation.

44. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions?

45. How much heat is produced by combustion of 125 g of methanol under standard state conditions?

46. How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions?

47. What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions?

48. When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions?

49. How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed?

\[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l)\hspace{20px}ΔH^\circ_{298}=\mathrm{−58\:kJ}\]

If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation?

50. A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74 °C to 27.93 °C. What is the heat capacity of the calorimeter and its contents?

51. Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, 6.00 kcal/mol) was used in household refrigerators. What mass of SO₂ must be evaporated to remove as much heat as evaporation of 1.00 kg of CCl₂F₂ (enthalpy of vaporization is 17.4 kJ/mol)?

The vaporization reactions for SO₂ and CCl₂F₂ are

\(\ce{SO2}(l)⟶\ce{SO2}(g)\) and \(\ce{CCl2F}(l)⟶\ce{CCl2F2}(g)\), respectively.

52. Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 °C, as the heat provided when 100 g of steam is cooled from 110 °C to 100 °C.

53. Which of the enthalpies of combustion in Table (Table 5.2) the table are also standard enthalpies of formation?

54. Does the standard enthalpy of formation of H₂O(g) differ from ΔH° for the reaction \(\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(g)\)?

55. Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO(s) to Hg(l) and O₂(g) under standard conditions?

56. How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn₃O₄(s) at standard state conditions?

57. How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Fe₂O₃(s) at standard state conditions?

58. The following sequence of reactions occurs in the commercial production of aqueous nitric acid:

\(\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(l)\hspace{20px}ΔH=\mathrm{−907\:kJ}\)

\(\ce{2NO}(g)+\ce{O2}(g)⟶\ce{2NO2}(g)\hspace{20px}ΔH=\mathrm{−113\:kJ}\)

\(\ce{3NO2}+\ce{H2O}(l)⟶\ce{2HNO2}(aq)+\ce{NO}(g)\hspace{20px}ΔH=\mathrm{−139\:kJ}\)

Determine the total energy change for the production of one mole of aqueous nitric acid by this process.

59. Both graphite and diamond burn.

\(\ce{C}(s,\:\ce{diamond})+\ce{O2}(g)⟶\ce{CO2}(g)\)

For the conversion of graphite to diamond:

\(\ce{C}(s,\:\ce{graphite})⟶\ce{C}(s,\:\ce{diamond})\hspace{20px}ΔH^\circ_{298}=\mathrm{1.90\:kJ}\)

Which produces more heat, the combustion of graphite or the combustion of diamond?

60. From the molar heats of formation in Appendix G, determine how much heat is required to evaporate one mole of water: \(\ce{H2O}(l)⟶\ce{H2O}(g)\)

61. Which produces more heat?

\(\ce{Os}(s)⟶\ce{2O2}(g)⟶\ce{OsO4}(s)\)

or

\(\ce{Os}(s)⟶\ce{2O2}(g)⟶\ce{OsO4}(g)\)

for the phase change \(\ce{OsO4}(s)⟶\ce{OsO4}(g)\hspace{20px}ΔH=\mathrm{56.4\:kJ}\)

62. Calculate \(ΔH^\circ_{298}\) for the process

\(\ce{Sb}(s)+\dfrac{5}{2}\ce{Cl2}(g)⟶\ce{SbCl5}(g)\)

from the following information:

\(\ce{Sb}(s)+\dfrac{3}{2}\ce{Cl2}(g)⟶\ce{SbCl3}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−314\:kJ}\)

\(\ce{SbCl3}(s)+\ce{Cl2}(g)⟶\ce{SbCl5}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−80\:kJ}\)

63. Calculate \(ΔH^\circ_{298}\) for the process \(\ce{Zn}(s)+\ce{S}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)\)

from the following information:

\(\ce{Zn}(s)+\ce{S}(s)⟶\ce{ZnS}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−206.0\:kJ}\)

\(\ce{ZnS}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−776.8\:kJ}\)

64. Calculate ΔH for the process

\(\ce{Hg2Cl2}(s)⟶\ce{2Hg}(l)+\ce{Cl2}(g)\)

from the following information:

\(\ce{Hg}(l)+\ce{Cl2}(g)⟶\ce{HgCl2}(s)\hspace{20px}ΔH=\mathrm{−224\:kJ}\)

\(\ce{Hg}(l)+\ce{HgCl2}(s)⟶\ce{Hg2Cl2}(s)\hspace{20px}ΔH=\mathrm{−41.2\:kJ}\)

65. Calculate \(ΔH^\circ_{298}\) for the process

\(\ce{Co3O4}(s)⟶\ce{3Co}(s)+\ce{2O2}(g)\)

from the following information:

\(\ce{Co}(s)+\dfrac{1}{2}\ce{O2}(g)⟶\ce{CoO}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−237.9\:kJ}\)

\(\ce{3Co}(s)+\ce{O2}(g)⟶\ce{Co3O4}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−177.5\:kJ}\)

66. Calculate the standard molar enthalpy of formation of NO(g) from the following data:

\(\ce{N2}(g)+\ce{2O2}⟶\ce{2NO2}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{66.4\:kJ}\)

\(\ce{2NO}(g)+\ce{O2}⟶\ce{2NO2}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−114.1\:kJ}\)

67. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:

1. \(\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g)\)
2. \(\ce{Si}(s)+\ce{2Cl2}(g)⟶\ce{SiCl4}(g)\)
3. \(\ce{Fe2O3}(s)+\ce{3H2}(g)⟶\ce{2Fe}(s)+\ce{3H2O}(l)\)
4. \(\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g)\)

68. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:

1. \(\ce{Si}(s)+\ce{2F2}(g)⟶\ce{SiF4}(g)\)
2. \(\ce{2C}(s)+\ce{2H2}(g)+\ce{O2}(g)⟶\ce{CH3CO2H}(l)\)
3. \(\ce{CH4}(g)+\ce{N2}(g)⟶\ce{HCN}(g)+\ce{NH3}(g)\);
4. \(\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g)\)

69. The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each.

1. \(\ce{2Ag2O}(s)⟶\ce{4Ag}(s)+\ce{O2}(g)\)
2. \(\ce{SnO}(s)+\ce{CO}(g)⟶\ce{Sn}(s)+\ce{CO2}(g)\)
3. \(\ce{Cr2O3}(s)+\ce{3H2}(g)⟶\ce{2Cr}(s)+\ce{3H2O}(l)\)
4. \(\ce{2Al}(s)+\ce{Fe2O3}(s)⟶\ce{Al2O3}(s)+\ce{2Fe}(s)\)

70. The decomposition of hydrogen peroxide, H₂O₂, has been used to provide thrust in the control jets of various space vehicles. Using the data in Appendix G, determine how much heat is produced by the decomposition of exactly 1 mole of H₂O₂ under standard conditions.

\(\ce{2H2O2}(l)⟶\ce{2H2O}(g)+\ce{O2}(g)\)

71. Calculate the enthalpy of combustion of propane, C₃H₈(g), for the formation of H₂O(g) and CO₂(g). The enthalpy of formation of propane is −104 kJ/mol.

72. Calculate the enthalpy of combustion of butane, C₄H₁₀(g) for the formation of H₂O(g) and CO₂(g). The enthalpy of formation of butane is −126 kJ/mol.

73. Both propane and butane are used as gaseous fuels. Which compound produces more heat per gram when burned?

74. The white pigment TiO₂ is prepared by the reaction of titanium tetrachloride, TiCl₄, with water vapor in the gas phase:

\(\ce{TiCl4}(g)+\ce{2H2O}(g)⟶\ce{TiO2}(s)+\ce{4HCl}(g)\).

How much heat is evolved in the production of exactly 1 mole of TiO₂(s) under standard state conditions?

75. Water gas, a mixture of H₂ and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon:

\(\ce{C}(s)+\ce{H2O}(g)⟶\ce{CO}(g)+\ce{H2}(g)\).

1. Assuming that coke has the same enthalpy of formation as graphite, calculate \(ΔH^\circ_{298}\) for this reaction.
2. Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst: \[\ce{2H2}(g)+\ce{CO}(g)⟶\ce{CH3OH}(g).\] Under the conditions of the reaction, methanol forms as a gas. Calculate \(ΔH^\circ_{298}\) for this reaction and for the condensation of gaseous methanol to liquid methanol.
3. Calculate the heat of combustion of 1 mole of liquid methanol to H₂O(g) and CO₂(g).

76. In the early days of automobiles, illumination at night was provided by burning acetylene, C₂H₂. Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC₂:

\(\ce{CaC2}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s)+\ce{C2H2}(g)\).

Calculate the standard enthalpy of the reaction. The \(ΔH^\circ_\ce{f}\) of CaC₂ is −15.14 kcal/mol.

77. From the data in Table (Table 5.2), determine which of the following fuels produces the greatest amount of heat per gram when burned under standard conditions: CO(g), CH₄(g), or C₂H₂(g).

78. The enthalpy of combustion of hard coal averages −35 kJ/g, that of gasoline, 1.28 × 10⁵ kJ/gal. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is 0.692 g/mL (the same as the density of isooctane).

79. Ethanol, C₂H₅OH, is used as a fuel for motor vehicles, particularly in Brazil.

1. Write the balanced equation for the combustion of ethanol to CO₂(g) and H₂O(g), and, using the data in Appendix G, calculate the enthalpy of combustion of 1 mole of ethanol.
2. The density of ethanol is 0.7893 g/mL. Calculate the enthalpy of combustion of exactly 1 L of ethanol.
3. Assuming that an automobile’s mileage is directly proportional to the heat of combustion of the fuel, calculate how much farther an automobile could be expected to travel on 1 L of gasoline than on 1 L of ethanol. Assume that gasoline has the heat of combustion and the density of n–octane, C₈H₁₈ (\(ΔH^\circ_\ce{f}=\mathrm{−208.4\:kJ/mol}\); density = 0.7025 g/mL).

80. Among the substances that react with oxygen and that have been considered as potential rocket fuels are diborane [B₂H₆, produces B₂O₃(s) and H₂O(g)], methane [CH₄, produces CO₂(g) and H₂O(g)], and hydrazine [N₂H₄, produces N₂(g) and H₂O(g)]. On the basis of the heat released by 1.00 g of each substance in its reaction with oxygen, which of these compounds offers the best possibility as a rocket fuel? The \(ΔH^\circ_\ce{f}\) of B₂H₆(g), CH₄(g), and N₂H₄(l) may be found in Appendix G.

81. How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions?

Ethylene, C₂H₂, a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About 80% of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst.

\(\ce{C2H4}(g)+\ce{H2O}(g)⟶\ce{C2H5OH}(l)\)

Using the data in the table in Appendix G, calculate ΔH° for the reaction.

82. The oxidation of the sugar glucose, C₆H₁₂O₆, is described by the following equation:

\(\ce{C6H12O6}(s)+\ce{6O2}(g)⟶\ce{6CO2}(g)+\ce{6H2O}(l)\hspace{20px}ΔH=\mathrm{−2816\:kJ}\)

The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body.

1. How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose?
2. How many Calories can be produced by the metabolism of 1.0 g of glucose?

83. Propane, C₃H₈, is a hydrocarbon that is commonly used as a fuel.

1. Write a balanced equation for the complete combustion of propane gas.
2. Calculate the volume of air at 25 °C and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent O₂ by volume. (Hint: we will see how to do this calculation in a later chapter on gases—for now use the information that 1.00 L of air at 25 °C and 1.00 atm contains 0.275 g of O₂ per liter.)
3. The heat of combustion of propane is −2,219.2 kJ/mol. Calculate the heat of formation, \(ΔH^\circ_\ce{f}\) of propane given that \(ΔH^\circ_\ce{f}\) of H₂O(l) = −285.8 kJ/mol and \(ΔH^\circ_\ce{f}\) of CO₂(g) = −393.5 kJ/mol.
4. Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water.

84. During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with 89% efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house).

1. Assume that natural gas is pure methane and determine the volume of natural gas in cubic feet that was required to heat the house. The average temperature of the natural gas was 56 °F; at this temperature and a pressure of 1 atm, natural gas has a density of 0.681 g/L.
2. How many gallons of LPG (liquefied petroleum gas) would be required to replace the natural gas used? Assume the LPG is liquid propane [C₃H₈: density, 0.5318 g/mL; enthalpy of combustion, 2219 kJ/mol for the formation of CO₂(g) and H₂O(l)] and the furnace used to burn the LPG has the same efficiency as the gas furnace.
3. What mass of carbon dioxide is produced by combustion of the methane used to heat the house?
4. What mass of water is produced by combustion of the methane used to heat the house?
5. What volume of air is required to provide the oxygen for the combustion of the methane used to heat the house? Air contains 23% oxygen by mass. The average density of air during the month was 1.22 g/L.
6. How many kilowatt–hours (1 kWh = 3.6 × 10⁶ J) of electricity would be required to provide the heat necessary to heat the house? Note electricity is 100% efficient in producing heat inside a house.
7. Although electricity is 100% efficient in producing heat inside a house, production and distribution of electricity is not 100% efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about 40%. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is 40%?

Solutions

S7.2: Energy Basics

S1. Temperature vs. Thermal Energy

Temperature is an intensive property that measures the average kinetic energy of the particles in a substance. While the average kinetic energy of the molecules in a match flame and a bonfire can be identical, the bonfire contains a vastly larger quantity of matter (mass). Heat (thermal energy) is an extensive property that depends directly on the total amount of matter present. Because the bonfire has millions of times more reacting particles than a single tiny match, it possesses and transfers a much greater total quantity of thermal energy, which is required to warm your body and the surrounding cold autumn air.

S2. Heat Capacity vs. Specific Heat

Heat Capacity (\(C\)): This is an extensive property defined as the quantity of heat required to change the temperature of an entire mass of an object by \(1^\circ\text{C}\) (or \(1\text{ K}\)). Its units are typically \(\text{J/}^\circ\text{C}\). For example, a swimming pool has a much higher heat capacity than a glass of water, even though they are made of the same substance.

Specific Heat (\(c_s\)): This is an intensive property defined as the quantity of heat required to change the temperature of exactly one gram of a substance by \(1^\circ\text{C}\) (or \(1\text{ K}\)). Its units are \(\text{J/g}\cdot^\circ\text{C}\). It remains constant for a specific material regardless of how much of it you have.

S3. Calculating Heat Capacity

a) 28.4 g of water (Specific heat of liquid water, \(c_s = 4.184 \text{ J/g}\cdot^\circ\text{C} = 1.00 \text{ cal/g}\cdot^\circ\text{C}\))

In Joules:

\[ C = m \cdot c_s = 28.4 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C} = 119 \text{ J/}^\circ\text{C} \]

In Calories:

\[ C = m \cdot c_s = 28.4 \text{ g} \times 1.00 \text{ cal/g}\cdot^\circ\text{C} = 28.4 \text{ cal/}^\circ\text{C} \]

b) 1.00 oz of lead (Specific heat of lead, \(c_s = 0.128 \text{ J/g}\cdot^\circ\text{C}\). Conversion: \(1 \text{ oz} = 28.35 \text{ g}\))

Convert mass to grams: \(1.00 \text{ oz} \times 28.35 \text{ g/oz} = 28.35 \text{ g}\)

In Joules:

\[ C = m \cdot c_s = 28.35 \text{ g} \times 0.128 \text{ J/g}\cdot^\circ\text{C} = 3.63 \text{ J/}^\circ\text{C} \]

In Calories (\(1 \text{ cal} = 4.184 \text{ J}\)):

\[ C = \frac{3.63 \text{ J/}^\circ\text{C}}{4.184 \text{ J/cal}} = 0.868 \text{ cal/}^\circ\text{C} \]

S4. Calculating Heat Capacity

a) 45.8 g of nitrogen gas (Specific heat of \(\text{N}_2\) gas at constant pressure, \(c_s = 1.040 \text{ J/g}\cdot^\circ\text{C}\))

In Joules:

\[ C = m \cdot c_s = 45.8 \text{ g} \times 1.040 \text{ J/g}\cdot^\circ\text{C} = 47.6 \text{ J/}^\circ\text{C} \]

In Calories:

\[ C = \frac{47.6 \text{ J/}^\circ\text{C}}{4.184 \text{ J/cal}} = 11.4 \text{ cal/}^\circ\text{C} \]

b) 1.00 pound of aluminum metal (Specific heat of aluminum, \(c_s = 0.897 \text{ J/g}\cdot^\circ\text{C}\). Conversion: \(1 \text{ lb} = 453.59 \text{ g}\))

Convert mass to grams: \(1.00 \text{ lb} = 453.59 \text{ g}\)

In Joules:

\[ C = m \cdot c_s = 453.59 \text{ g} \times 0.897 \text{ J/g}\cdot^\circ\text{C} = 407 \text{ J/}^\circ\text{C} \]

In Calories:

\[ C = \frac{407 \text{ J/}^\circ\text{C}}{4.184 \text{ J/cal}} = 97.3 \text{ cal/}^\circ\text{C} \]

S5. Heating an Iron Block

Temperature change (\(\Delta T\)):** \[ \Delta T = 1535^\circ\text{C} - 25^\circ\text{C} = 1510^\circ\text{C} \]

In Joules:

\[ q = m \cdot c_s \cdot \Delta T \]

\[ q = 75.0 \text{ g} \times 0.449 \text{ J/g}\cdot^\circ\text{C} \times 1510^\circ\text{C} = 50,849 \text{ J} \implies 5.08 \times 10^4 \text{ J} \text{ (or } 50.8 \text{ kJ)} \]

In Calories:

\[ q = \frac{50,849 \text{ J}}{4.184 \text{ J/cal}} = 12,153 \text{ cal} \implies 1.22 \times 10^4 \text{ cal} \text{ (or } 12.2 \text{ kcal)} \]

S6. Heating an Ice Cube

Note: Use the specific heat capacity of solid ice, \(c_s = 2.093 \text{ J/g}\cdot^\circ\text{C} = 0.500 \text{ cal/g}\cdot^\circ\text{C}\))

Temperature change (\(\Delta T\)): \[ \Delta T = -1.0^\circ\text{C} - (-23.0^\circ\text{C}) = 22.0^\circ\text{C} \]

In Joules:

\[ q = m \cdot c_s \cdot \Delta T \]

\[ q = 28.4 \text{ g} \times 2.093 \text{ J/g}\cdot^\circ\text{C} \times 22.0^\circ\text{C} = 1,308 \text{ J} \implies 1.31 \times 10^3 \text{ J} \text{ (or } 1.31 \text{ kJ)} \]

In Calories:

\[ q = m \cdot c_s \cdot \Delta T = 28.4 \text{ g} \times 0.500 \text{ cal/g}\cdot^\circ\text{C} \times 22.0^\circ\text{C} = 312.4 \text{ cal} \implies 312 \text{ cal} \]

S7. Temperature Increase of Water

Convert energy units: \(36.5 \text{ kJ} = 36,500 \text{ J}\)

Rearrange the heat formula (\(q = m \cdot c_s \cdot \Delta T\)) to solve for \(\Delta T\):

\[ \Delta T = \frac{q}{m \cdot c_s} \]

\[ \Delta T = \frac{36,500 \text{ J}}{275 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C}} \]

\[ \Delta T = \frac{36,500}{1150.6} = 31.7^\circ\text{C} \]

The temperature of the water would increase by \(31.7^\circ\text{C}\).

S8. Temperature Increase of Water

Convert energy units: \(14.5 \text{ kJ} = 14,500 \text{ J}\)

Solve for temperature change (\(\Delta T\)):

\[ \Delta T = \frac{q}{m \cdot c_s} \]

\[ \Delta T = \frac{14,500 \text{ J}}{485 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C}} \]

\[ \Delta T = \frac{14,500}{2029.24} = 7.15^\circ\text{C} \]

The temperature of the water would increase by \(7.15^\circ\text{C}\).

S9. Identifying an Unknown Substance

Step 1: Calculate the temperature change (\(\Delta T\))

\[ \Delta T = 89.6^\circ\text{C} - 23.2^\circ\text{C} = 66.4^\circ\text{C} \]

Step 2: Calculate the specific heat (\(c_s\)) using the rearranged heat equation

\[ q = m \cdot c_s \cdot \Delta T \implies c_s = \frac{q}{m \cdot \Delta T} \]

\[ c_s = \frac{2110 \text{ J}}{44.7 \text{ g} \times 66.4^\circ\text{C}} = \frac{2110}{2968.08} = 0.711 \text{ J/g}\cdot^\circ\text{C} \]

Identity: Looking at Table 5.1, a value of \(0.711 \text{ J/g}\cdot^\circ\text{C}\) matches silicon (\(\text{Si}(s)\)), which has a listed specific heat of \(0.712 \text{ J/g}\cdot^\circ\text{C}\).

S10. Identifying an Unknown Substance

Step 1: Calculate the temperature change (\(\Delta T\))

\[ \Delta T = 68.9^\circ\text{C} - 19.3^\circ\text{C} = 49.6^\circ\text{C} \]

Step 2: Calculate the specific heat (\(c_s\))

\[ c_s = \frac{q}{m \cdot \Delta T} \]

\[ c_s = \frac{8460 \text{ J}}{437.2 \text{ g} \times 49.6^\circ\text{C}} = \frac{8460}{21685.12} = 0.390 \text{ J/g}\cdot^\circ\text{C} \]

Identity: Looking at Table 5.1, a value of \(0.390 \text{ J/g}\cdot^\circ\text{C}\) is closest to copper (\(\text{Cu}(s)\)), which has a listed specific heat of \(0.385 \text{ J/g}\cdot^\circ\text{C}\).

S11. Aluminum Kettle Calculatiuons

Note: From Table 5.1, the specific heat of aluminum \(c_{s,\text{Al}} = 0.897 \text{ J/g}\cdot^\circ\text{C}\))

a) Heat Capacity (\(C_{\text{kettle}}\)):

Convert the mass of the kettle to grams: \(1.05 \text{ kg} = 1050 \text{ g}\).

\[ C_{\text{kettle}} = m \cdot c_{s,\text{Al}} = 1050 \text{ g} \times 0.897 \text{ J/g}\cdot^\circ\text{C} = 941.85 \text{ J/}^\circ\text{C} \implies 942 \text{ J/}^\circ\text{C} \]

b) Heat required for the empty kettle (\(q_{\text{kettle}}\)):

\[ \Delta T = 99.0^\circ\text{C} - 23.0^\circ\text{C} = 76.0^\circ\text{C} \]

\[ q_{\text{kettle}} = C_{\text{kettle}} \cdot \Delta T = 941.85 \text{ J/}^\circ\text{C} \times 76.0^\circ\text{C} = 71,581 \text{ J} \implies 7.16 \times 10^4 \text{ J} \text{ (or } 71.6 \text{ kJ)} \]

c) Heat required for the kettle filled with water (\(q_{\text{total}}\)):

Step 1: Find the mass of the water (\(m_{\text{water}}\))

\[ m_{\text{water}} = \text{Volume} \times \text{Density} = 1250 \text{ mL} \times 0.997 \text{ g/mL} = 1246.25 \text{ g} \]

Step 2: Find the heat required to warm the water (\(q_{\text{water}}\))

\[ q_{\text{water}} = m_{\text{water}} \cdot c_{s,\text{water}} \cdot \Delta T \]

\[ q_{\text{water}} = 1246.25 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C} \times 76.0^\circ\text{C} = 396,327 \text{ J} \]

Step 3: Sum the heat values together

\[ q_{\text{total}} = q_{\text{kettle}} + q_{\text{water}} = 71,581 \text{ J} + 396,327 \text{ J} = 467,908 \text{ J} \implies 4.68 \times 10^5 \text{ J} \text{ (or } 468 \text{ kJ)} \]

S12. Waterbed Electrical Energy Estimate

Step 1: Convert the temperature change from Fahrenheit to Celsius

\(T_{\text{initial}} = (85^\circ\text{F} - 32) \times \frac{5}{9} = 29.44^\circ\text{C}\)

\(T_{\text{final}} = (72^\circ\text{F} - 32) \times \frac{5}{9} = 22.22^\circ\text{C}\)

\(\Delta T = 29.44^\circ\text{C} - 22.22^\circ\text{C} = 7.22^\circ\text{C}\)

Step 2: Find the mass of the water

\[ 892 \text{ L} = 892,000 \text{ mL} \times 1.0 \text{ g/mL} = 892,000 \text{ g} \]

Step 3: Calculate the thermal energy lost in Joules

\[ q = m \cdot c_s \cdot \Delta T \]

\[ q = 892,000 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C} \times 7.22^\circ\text{C} = 2.694 \times 10^7 \text{ J} \]

Step 4: Convert Joules to kilowatt-hours (kWh)

\[ \text{Electrical Energy} = \frac{2.694 \times 10^7 \text{ J}}{3.6 \times 10^6 \text{ J/kWh}} = 7.48 \text{ kWh} \implies 7.5 \text{ kWh} \]

Assumptions & Error Analysis:

1. Assumption: We assumed the heating element behaves with \(100\%\) thermodynamic efficiency (all electrical energy converts perfectly to heat absorbed by the water).

Impact: In reality, some energy is lost directly to the floor or surrounding frame. This assumption yields a negative error (underestimates the true electrical requirement).

2. Assumption: We assumed no heat was absorbed or lost by the structural mattress vinyl, blankets, or bed frame materials.

Impact: The surrounding materials have their own heat capacities. Ignoring them yields a negative error.

3. Assumption: We assumed the specific heat capacity of water stays perfectly uniform at \(4.184 \text{ J/g}\cdot^\circ\text{C}\) across this temperature window.

Impact: The deviation of water's specific heat between \(22^\circ\text{C}\) and \(29^\circ\text{C}\) is extremely marginal, yielding a negligible error.

S7.3: Calorimetry

S13. Thermal Energy Loss: Volume and Rates

The second student is correct. The 2-L bottle of water lost more heat because it contains a larger mass of water undergoing the exact same temperature change.

Analysis of Each Student's Answer:

Student 1 is incorrect: The Error: This student is confusing temperature (an intensive property measuring average kinetic energy) with heat (an extensive property measuring total thermal energy transferred). While the temperature change (\(\Delta T\)) is identical for both samples, the 2-L bottle contains four times the mass of the 500-mL bottle (\(2\text{ L vs. } 0.5\text{ L}\)). Because heat transfer is directly proportional to mass (\(q = m \cdot c_s \cdot \Delta T\)), the larger volume must release substantially more total energy to achieve that same drop in temperature.

Student 2 is correct: The Reasoning: This student correctly recognizes that heat is an extensive property. Since both bottles start at the same initial temperature and cool to the same final refrigerator temperature, their temperature change (\(\Delta T\)) is identical. Because the 2-L bottle has a much larger mass (\(m\)), it requires a much larger loss of thermal energy (\(q\)) to cool down.

Student 3 is incorrect: The Error: This student is confusing the rate of heat transfer with the total quantity of heat transferred. The 500-mL bottle cools faster simply because it has a larger surface-area-to-volume ratio, allowing heat to escape to the refrigerator more rapidly per unit of mass. However, a faster rate of cooling over a short time does not equate to a larger cumulative quantity of energy lost.

Student 4 is incorrect: The Error: While it is true that we do not know the exact numerical values of the initial and final temperatures, we are explicitly told that both bottles started at the same temperature (room temperature) and ended at the same temperature (refrigerator temperature). Because the terms are held constant relative to each other, \(\Delta T\) is identical for both. Therefore, we have more than enough qualitative variables to determine that the larger mass lost more heat.

S14. Impact of Poor Insulation on an Exothermic Reaction (Example 5.5)

The amount of heat measured (calculated from the experimental temperature change) would be lesser than the true value.

Explanation: Example 5.5 describes an exothermic reaction that releases heat into the aqueous solution, causing the temperature of the thermometer to rise (\(22.0^\circ\text{C} \rightarrow 28.9^\circ\text{C}\)).

If the calorimeter is a poorer insulator, a significant portion of the heat produced by the chemical reaction will escape into the outside surroundings before it can be fully absorbed by the water. As a result, the water will not reach its maximum possible temperature, yielding a lower experimental \(\Delta T\). Because our calculation relies directly on this temperature change (\(q_{\text{solution}} = m \cdot c_s \cdot \Delta T\)), a smaller \(\Delta T\) will cause the calculated amount of heat produced to appear lower than it actually is.

S15. Impact of Poor Insulation on an Endothermic Dissolution (Example 5.6)

The amount of heat absorbed by the dissolution would appear lesser (underestimated) than the true value.

Explanation: Example 5.6 describes an endothermic dissolution (an instant cold pack) where the chemical process absorbs heat *from* the water, causing the temperature of the solution to drop (\(24.9^\circ\text{C} \rightarrow 20.3^\circ\text{C}\)).

If the calorimeter is a poor insulator, environmental heat from the warmer room air will leak *into* the cold cup during the experiment. This incoming environmental heat partially warms the water, preventing the temperature from dropping as low as it ideally should have. This results in a smaller observed temperature drop (\(\Delta T\)). Because the calculation relies directly on this value (\(q_{\text{soln}} = m \cdot c_s \cdot \Delta T\)), a smaller temperature change causes the process to falsely appear as if it absorbed less heat than it actually did.

S16. Accounting for Calorimeter Heat Capacity (Example 5.6)

The total amount of heat absorbed by the dissolution would appear greater than the simplified calculation.

Explanation: In the simplified calculation provided in Example 5.6, it is assumed that only the water changes temperature. However, the physical walls of the calorimeter (the coffee cup itself) and the thermometer are also in direct contact with the solution, meaning they cool down alongside the water from \(24.9^\circ\text{C}\) to \(20.3^\circ\text{C}\).

If we account for the heat capacity of the calorimeter (\(C_{\text{cal}}\)), the total heat of the process must include the thermal energy pulled from the cup itself:

\[ q_{\text{rxn}} = -(q_{\text{soln}} + q_{\text{cal}}) \]

Where \(q_{\text{cal}} = C_{\text{cal}} \cdot \Delta T\). Because the chemical system is pulling heat out of both the water *and* the calorimeter hardware to drive the dissolution, taking the calorimeter into account reveals that a larger total quantity of thermal energy was absorbed.

S17. Mixing Water and Coffee (Thermal Equilibrium)

Because the coffee and water are mixed in an isolated system, the heat lost by the hot coffee must equal the heat gained by the cold water:

\[ q_{\text{lost}} = -q_{\text{gained}} \implies m_{\text{coffee}} \cdot c_s \cdot \Delta T_{\text{coffee}} = -(m_{\text{water}} \cdot c_s \cdot \Delta T_{\text{water}}) \]

Since both substances have the same specific heat (\(c_s\)), it cancels out from both sides:

\[ m_{\text{coffee}} \cdot (T_f - T_{i,\text{coffee}}) = -m_{\text{water}} \cdot (T_f - T_{i,\text{water}}) \]

Step 1: Substitute the known values

Given density = \(1.00 \text{ g/mL}\), the mass of 180 mL of coffee is \(180 \text{ g}\).

\(T_f = 60^\circ\text{C}\)

\(T_{i,\text{coffee}} = 95^\circ\text{C}\)

\(T_{i,\text{water}} = 23^\circ\text{C}\)

\[ 180 \text{ g} \times (60^\circ\text{C} - 95^\circ\text{C}) = -m_{\text{water}} \times (60^\circ\text{C} - 23^\circ\text{C}) \]

\[ 180 \times (-35) = -m_{\text{water}} \times (37) \]

\[ -6300 = -37 \cdot m_{\text{water}} \]

Step 2: Solve for mass and volume of water

\[ m_{\text{water}} = \frac{-6300}{-37} = 170.27 \text{ g} \]

Since the density of water is \(1.00 \text{ g/mL}\), the required volume is 170 mL (rounded to two significant figures).

S18. Cooling Coffee with a Silver Spoon

Step 1: Set up the thermal equilibrium equation**

\[ q_{\text{coffee}} = -q_{\text{spoon}} \]

\[ m_{\text{coffee}} \cdot c_{s,\text{water}} \cdot (T_f - T_{i,\text{coffee}}) = -[m_{\text{spoon}} \cdot c_{s,\text{Ag}} \cdot (T_f - T_{i,\text{spoon}})] \]

Step 2: Substitute values and expand

\[ 180 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C} \times (T_f - 95) = -[45 \text{ g} \times 0.24 \text{ J/g}\cdot^\circ\text{C} \times (T_f - 25)] \]

\[ 753.12 \times (T_f - 95) = -10.8 \times (T_f - 25) \]

\[ 753.12 T_f - 71,546.4 = -10.8 T_f + 270 \]

Step 3: Solve for the final temperature (\(T_f\))

\[ 753.12 T_f + 10.8 T_f = 71,546.4 + 270 \]

\[ 763.92 T_f = 71,816.4 \]

\[ T_f = \frac{71,816.4}{763.92} = 94.01^\circ\text{C} \]

Step 4: Calculate the temperature reduction (\(\Delta T\))

\[ \text{Reduction} = T_{i,\text{coffee}} - T_f = 95^\circ\text{C} - 94.01^\circ\text{C} = 0.99^\circ\text{C} \]

The temperature of the coffee will be reduced by \(1.0^\circ\text{C}\).

S19. Cooling Coffee with an Aluminum Spoon

a) Finding the final temperature:**

\[ q_{\text{coffee}} = -q_{\text{spoon}} \]

\[ 180 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C} \times (T_f - 85) = -[45 \text{ g} \times 0.88 \text{ J/g}\cdot^\circ\text{C} \times (T_f - 24)] \]

\[ 753.12 \times (T_f - 85) = -39.6 \times (T_f - 24) \]

\[ 753.12 T_f - 64,015.2 = -39.6 T_f + 950.4 \]

\[ 792.72 T_f = 64,965.6 \]

\[ T_f = \frac{64,965.6}{792.72} = 81.95^\circ\text{C} \implies 82^\circ\text{C} \]

b) Why 88 °C is physically impossible:

An answer of \(88^\circ\text{C}\) is completely incorrect because it violates the Second Law of Thermodynamics. Heat naturally transfers spontaneously from a hotter object to a colder object until thermal equilibrium is established. Because the initial coffee temperature was \(85^\circ\text{C}\) and the spoon was \(24^\circ\text{C}\), the final equilibrium temperature must fall somewhere within the range between \(24^\circ\text{C}\) and \(85^\circ\text{C}\). The coffee cannot spontaneously gain thermal energy from a cold spoon to become hotter than its starting temperature.

S20. Automobile Radiator Heat Transfer

Step 1: Convert temperature changes to Celsius

Instead of converting the absolute endpoints, we can find the difference in Fahrenheit (\(\Delta T_{\text{F}}\)) and convert the scale magnitude directly:

\[ \Delta T_{\text{F}} = 240^\circ\text{F} - 175^\circ\text{F} = 65^\circ\text{F} \]

\[ \Delta T_{\text{C}} = 65^\circ\text{F} \times \frac{5^\circ\text{C}}{9^\circ\text{F}} = 36.11^\circ\text{C} \]

Step 2: Convert 1 gallon of water to mass in grams

\(1 \text{ gallon} = 3.785 \text{ L} = 3785 \text{ mL}\)

Using the standard density of water (\(1.00 \text{ g/mL}\)):

\[ m = 3785 \text{ g} \]

Step 3: Calculate the heat transferred (\(q\))

\[ q = m \cdot c_s \cdot \Delta T \]

\[ q = 3785 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C} \times 36.11^\circ\text{C} \]

\[ q = 5.72 \times 10^5 \text{ J} \text{ (or } 572 \text{ kJ)} \]

The amount of heat transferred to the surroundings is \(5.7 \times 10^5 \text{ J}\).

S21. Identifying Specific Heat of a Metal

Relevance of Figure 5.12 for all questions:

Key Features:

Constant Pressure ($P$): Because the styrofoam cup assembly is enclosed by a loose-fitting cover rather than a rigid airtight seal, the pressure inside remains equal to the atmospheric pressure of the room. Therefore, the measured heat flow is equal to the enthalpy change:

\[ q_{\text{rxn}} = \Delta H \]

Nested Polystyrene Cups: Using two nested styrofoam cups creates an insulating barrier of trapped air pockets. This minimizes thermal exchange with the room environment, ensuring that almost all heat released or absorbed stays within the liquid solution.

The Role of the Stirrer: Continuous, gentle stirring is critical to ensure that the reaction mixture maintains a completely uniform temperature distribution, allowing the thermometer bulb to record the true maximum or minimum equilibrium temperature.

Problem Solution:

Note: Use the standard specific heat of water, \(c_{s,\text{water}} = 4.184 \text{ J/g}\cdot^\circ\text{C}\)

Step 1: Calculate the heat absorbed by the water (\(q_{\text{water}}\))

\[ \Delta T_{\text{water}} = 24.6^\circ\text{C} - 22.0^\circ\text{C} = 2.6^\circ\text{C} \]

\[ q_{\text{water}} = m \cdot c_s \cdot \Delta T = 100 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C} \times 2.6^\circ\text{C} = 1087.84 \text{ J} \]

The heat given up by the metal is equal in magnitude but opposite in sign:

\[ q_{\text{metal}} = -1087.84 \text{ J} \implies 1.1 \times 10^3 \text{ J} \text{ (or } 1.1 \text{ kJ released)} \]

Step 2: Calculate the specific heat of the metal (\(c_{s,\text{metal}}\))

\[ \Delta T_{\text{metal}} = 24.6^\circ\text{C} - 80.0^\circ\text{C} = -55.4^\circ\text{C} \]

\[ q_{\text{metal}} = m \cdot c_{s,\text{metal}} \cdot \Delta T_{\text{metal}} \]

\[ -1087.84 \text{ J} = 70.0 \text{ g} \times c_{s,\text{metal}} \times (-55.4^\circ\text{C}) \]

\[ -1087.84 = -3878 \cdot c_{s,\text{metal}} \]

\[ c_{s,\text{metal}} = \frac{-1087.84}{-3878} = 0.28 \text{ J/g}\cdot^\circ\text{C} \]

S22. Final Temperature After Heat Trapping

Step 1: Rearrange the heat formula to find the temperature change (\(\Delta T\))

Convert energy to Joules: \(1.506 \text{ kJ} = 1506 \text{ J}\).

\[ q = m \cdot c_s \cdot \Delta T \implies \Delta T = \frac{q}{m \cdot c_s} \]

\[ \Delta T = \frac{1506 \text{ J}}{30.0 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C}} = \frac{1506}{125.52} = 12.0^\circ\text{C} \]

Step 2: Calculate the final temperature (\(T_{\text{final}}\))

\[ T_{\text{final}} = T_{\text{initial}} + \Delta T \]

\[ T_{\text{final}} = 26.5^\circ\text{C} + 12.0^\circ\text{C} = 38.5^\circ\text{C} \]

S23. Dissolution Energy of \(\text{KCl}\)

Step 1: Determine the total mass of the final solution

\[ m_{\text{solution}} = 0.500 \text{ g (solute)} + 50.0 \text{ g (solvent)} = 50.500 \text{ g} \]

Step 2: Calculate the heat change of the solution (\(q_{\text{solution}}\))

Because the temperature decreases, \(\Delta T = -1.05^\circ\text{C}\).

\[ q_{\text{solution}} = m \cdot c_s \cdot \Delta T \]

\[ q_{\text{solution}} = 50.500 \text{ g} \times 4.18 \text{ J/g}\cdot^\circ\text{C} \times (-1.05^\circ\text{C}) = -221.64 \text{ J} \]

Step 3: Determine the heat of the reaction/dissolution (\(q_{\text{rxn}}\))

\[ q_{\text{rxn}} = -q_{\text{solution}} = +221.64 \text{ J} \implies 222 \text{ J} \]

Thermodynamic Classification: Because the arithmetic sign of \(q_{\text{rxn}}\) is positive (and the surrounding water temperature decreased as a result of losing its thermal energy to the chemical bonds), the process is endothermic.

S24. Dissolution Energy of \(\text{CaCl}_2\)

Step 1: Calculate total mass and temperature change

Total solution mass: \(m = 3.0 \text{ g} + 150.0 \text{ g} = 153.0 \text{ g}\)

Temperature change: \(\Delta T = 25.8^\circ\text{C} - 22.4^\circ\text{C} = 3.4^\circ\text{C}\)

Step 2: Calculate the heat of the solution

\[ q_{\text{solution}} = 153.0 \text{ g} \times 4.18 \text{ J/g}\cdot^\circ\text{C} \times 3.4^\circ\text{C} = 2174.3 \text{ J} \]

Step 3: Determine the heat involved in the dissolution

\[ q_{\text{rxn}} = -q_{\text{solution}} = -2174.3 \text{ J} \implies -2.2 \times 10^3 \text{ J} \text{ (or } -2.2 \text{ kJ)} \]

Thermodynamic Classification: Because the system released energy into the water causing the temperature to rise (yielding a negative sign for \(q_{\text{rxn}}\)), the reaction is exothermic.

S25. Precipitation Heat of Silver Chloride

Step 1: Calculate total combined mass and temperature change

Combined mass: \(m = 50.0 \text{ g} + 100.0 \text{ g} = 150.0 \text{ g}\)

Temperature change: \(\Delta T = 25.2^\circ\text{C} - 24.1^\circ\text{C} = 1.1^\circ\text{C}\)

Step 2: Calculate the heat absorbed by the solution

\[ q_{\text{solution}} = m \cdot c_s \cdot \Delta T \]

\[ q_{\text{solution}} = 150.0 \text{ g} \times 4.20 \text{ J/g}\cdot^\circ\text{C} \times 1.1^\circ\text{C} = 693 \text{ J} \]

Conclusion: The heat produced by the chemical precipitation reaction is 690 J (rounded to two significant figures to match the temperature limits).

S26. Heat Absorbed by Endothermic Coupling

Step 1: Calculate the total combined mass of the solution contents

\[ m_{\text{total}} = 3.15 \text{ g} + 1.52 \text{ g} + 100 \text{ g} = 104.67 \text{ g} \]

Step 2: Calculate the heat flow of the surroundings (\(q_{\text{solution}}\))

Since the temperature falls, \(\Delta T = -3.1^\circ\text{C}\).

\[ q_{\text{solution}} = m_{\text{total}} \cdot c_s \cdot \Delta T \]

\[ q_{\text{solution}} = 104.67 \text{ g} \times 4.20 \text{ J/g}\cdot^\circ\text{C} \times (-3.1^\circ\text{C}) = -1362.8 \text{ J} \]

Step 3: Determine the heat absorbed by the reaction (\(q_{\text{rxn}}\))

\[ q_{\text{rxn}} = -q_{\text{solution}} = +1362.8 \text{ J} \]

Rounding to two significant figures to match the temperature measurement: The heat absorbed by the reaction is \(1.4 \times 10^3 \text{ J}\) (or 1.4 kJ).

S27. Scaling an Exothermic Neutralization

The temperature would increase by the same amount, 6.9 °C (assuming the heat capacity of the calorimeter hardware itself is negligible).

Explanation: Doubling the volumes of both the acid and base reactants doubles the absolute amount of moles participating in the neutralization reaction, which doubles the total quantity of heat produced (\(q\)).

However, doubling the volumes also doubles the mass (\(m\)) of the final aqueous solution that must absorb this heat. According to the heat equation:

\[ \Delta T = \frac{q}{m \cdot c_s} \]

Since both the heat generated (\(q\)) and the absorbing mass (\(m\)) increase by a factor of 2 simultaneously, the ratio remains completely unchanged, meaning the temperature change (\(\Delta T\)) stays exactly the same.

S28. Altering Solvent Mass in Dissolution

The temperature would decrease by approximately 2.4 °C.

Explanation:

From the baseline calculations in Example 5.6, dissolving 3.21 g of \(\text{NH}_4\text{NO}_3\) absorbs exactly \(1.0 \times 10^3 \text{ J}\) of heat from its surroundings (\(q_{\text{solution}} = -1.0 \times 10^3 \text{ J}\)).

New total solution mass: \(m = 3.21 \text{ g} + 100.0 \text{ g} = 103.21 \text{ g}\)

Specific heat of solution: \(c_s = 4.184 \text{ J/g}\cdot^\circ\text{C}\)

Solve for the new \(\Delta T\):

\[ \Delta T = \frac{q_{\text{solution}}}{m \cdot c_s} \]

\[ \Delta T = \frac{-1.0 \times 10^3 \text{ J}}{103.21 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C}} = \frac{-1000}{431.83} = -2.31^\circ\text{C} \]

The temperature drops by 2.3 °C (or 2.4 °C depending on rounding thresholds from Example 5.6's precise joule figures).

S29. Combustion of Fructose in a Bomb Calorimeter

A bomb calorimeter operates at a constant volume. The heat absorbed by the calorimeter hardware and its internal water jacket combined is given directly by its total heat capacity (\(C_{\text{cal}}\)):

\[ q_{\text{cal}} = C_{\text{cal}} \cdot \Delta T \]

\[ q_{\text{cal}} = 9.90 \text{ kJ/}^\circ\text{C} \times 1.58^\circ\text{C} = 15.64 \text{ kJ} \]

By the law of conservation of energy, the heat released by the combustion system is the negative of the heat absorbed by the instrument:

\[ q_{\text{combustion}} = -q_{\text{cal}} = -15.6 \text{ kJ} \]

S30. Combustion of Trinitrotoluene (TNT)

Step 1: Calculate the temperature change (\(\Delta T\))

\[ \Delta T = 26.9^\circ\text{C} - 23.4^\circ\text{C} = 3.5^\circ\text{C} \]

Step 2: Calculate heat absorbed by the internal water jacket (\(q_{\text{water}}\))

Given 675 mL of water matches a mass of 675 g (density = 1.00 g/mL):

\[ q_{\text{water}} = m \cdot c_s \cdot \Delta T \]

\[ q_{\text{water}} = 675 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C} \times 3.5^\circ\text{C} = 9884.7 \text{ J} \]

Step 3: Calculate heat absorbed by the dry calorimeter hardware (\(q_{\text{hardware}}\))

\[ q_{\text{hardware}} = C_{\text{cal}} \cdot \Delta T \]

\[ q_{\text{hardware}} = 534 \text{ J/}^\circ\text{C} \times 3.5^\circ\text{C} = 1869 \text{ J} \]

Step 4: Sum to find total heat produced

\[ q_{\text{total produced}} = q_{\text{water}} + q_{\text{hardware}} \]

\[ q_{\text{total produced}} = 9884.7 \text{ J} + 1869 \text{ J} = 11,753.7 \text{ J} \implies 1.18 \times 10^4 \text{ J} \text{ (or 11.8 kJ)} \]

S31. Mass Scaling for Industrial Coal Combustion

Step 1: Calculate heat produced per single gram of coal

\[ q_{\text{cal}} = C_{\text{cal}} \cdot \Delta T \]

\[ q_{\text{cal}} = 21.6 \text{ kJ/}^\circ\text{C} \times 1.48^\circ\text{C} = 31.968 \text{ kJ per gram} \]

Step 2: Convert one ton of coal to grams

\(1 \text{ pound (lb)} = 453.592 \text{ g}\)

\(1 \text{ ton} = 2000 \text{ lbs}\)

\[ m_{\text{ton}} = 2000 \text{ lbs} \times 453.592 \text{ g/lb} = 907,184 \text{ g} \]

Step 3: Scale the heat output to one full ton

\[ \text{Total Heat} = 31.968 \text{ kJ/g} \times 907,184 \text{ g} = 2.90 \times 10^7 \text{ kJ} \]

The total heat produced by the combustion of a ton of coal is \(2.90 \times 10^7 \text{ kJ}\) (or \(2.90 \times 10^{10} \text{ J}\)).

S32. Dietary Fat Calorie Percentage

Step 1: Calculate the total Calories supplied by the fat

\[ \text{Calories from fat} = 65 \text{ g} \times 9.1 \text{ Calories/g} = 591.5 \text{ Calories} \]

Step 2: Calculate the percentage of the daily diet

\[ \text{Percentage} = \left( \frac{591.5 \text{ Calories}}{2000 \text{ Calories}} \right) \times 100\% = 29.575\% \]

Rounding to two significant figures to match the limit of the daily diet and mass inputs: The fat supplies 30% of the total calories in the diet.

S33. Mass of a Teaspoon of Sugar

Using dimensional analysis to rearrange the relationship:

\[ \text{Mass} = \frac{\text{Total Calories}}{\text{Calorie Density}} \]

\[ \text{Mass} = \frac{16 \text{ Calories}}{4.1 \text{ Calories/g}} = 3.902 \text{ g} \]

Rounding to two significant figures: The mass of one teaspoon of sucrose is 3.9 g.

S34. Maximum Carbohydrate in Diet Soda

If the can contains less than 1 Calorie, we evaluate the upper threshold limit of exactly 1 Calorie:

\[ \text{Maximum Mass} = \frac{1 \text{ Calorie}}{4.1 \text{ Calories/g}} = 0.2439 \text{ g} \]

Rounding to one significant figure based on the 1 Calorie parameter: The maximum mass of carbohydrate is 0.2 g.

S35. Fat Storage from Excess Calories

Step 1: Calculate the mass in grams

\[ \text{Mass in grams} = \frac{1100 \text{ Calories}}{9.1 \text{ Calories/g}} = 120.88 \text{ g} \]

Rounding to two significant figures matches 120 g.

Step 2: Convert the mass to pounds (\(1 \text{ lb} = 453.592 \text{ g}\))

\[ \text{Mass in pounds} = \frac{120.88 \text{ g}}{453.592 \text{ g/lb}} = 0.2665 \text{ lb} \]

Rounding to two significant figures matches 0.27 lb.

S36. Total Calories in a Cereal Serving

Sum the Calorie contribution of each individual macronutrient component:

From Protein: \(3 \text{ g} \times 4.1 \text{ Calories/g} = 12.3 \text{ Calories}\\)

From Carbohydrates: \(18 \text{ g} \times 4.1 \text{ Calories/g} = 73.8 \text{ Calories}\\)

From Fat: \(6 \text{ g} \times 9.1 \text{ Calories/g} = 54.6 \text{ Calories}\\)

\[ \text{Total Calories} = 12.3 + 73.8 + 54.6 = 140.7 \text{ Calories} \]

Rounding to the nearest whole number (or tracking to the units place for addition sig-fig alignment): The total energy content per serving is 141 Calories (or \(1.4 \times 10^2 \text{ Calories}\)).

S37. Cost-Efficiency Comparison: Cereal vs. Isooctane

Part 1: Energy per Dollar for the Breakfast Cereal

Step 1: Find total Calories in the box

The box holds 32 ounces, and each 1.0-ounce serving provides 130 Calories:

\[ 32 \text{ servings} \times 130 \text{ Calories/serving} = 4160 \text{ Calories} \]

Step 2: Convert Calories to Kilojoules (\(1 \text{ Calorie} = 4.184 \text{ kJ}\))

\[ 4160 \text{ Calories} \times 4.184 \text{ kJ/Calorie} = 17,405.4 \text{ kJ} \]

Step 3: Calculate Kilojoules per Dollar

\[ \text{Energy density per dollar} = \frac{17,405.4 \text{ kJ}}{\$4.23} = 4115 \text{ kJ/\$} \]

Part 2: Energy per Dollar for the Isooctane Gas

Step 1: Find the mass of 1.0 Liter (1000 mL) of isooctane

\[ m = 1000 \text{ mL} \times 0.6919 \text{ g/mL} = 691.9 \text{ g} \]

Step 2: Calculate the energy from chemical combustion

The standard enthalpy of combustion (\(\Delta H_c^\circ\)) for liquid isooctane (\(\text{C}_8\text{H}_{18}\)) is approximately \(-5461 \text{ kJ/mol}\). Its molar mass is \(114.23 \text{ g/mol}\).

\[ \text{Moles of isooctane} = \frac{691.9 \text{ g}}{114.23 \text{ g/mol}} = 6.057 \text{ mol} \]

\[ \text{Total heat produced} = 6.057 \text{ mol} \times 5461 \text{ kJ/mol} = 33,077.3 \text{ kJ} \]

Step 3: Calculate Kilojoules per Dollar

\[ \text{Energy density per dollar} = \frac{33,077.3 \text{ kJ}}{\$0.45} = 73,505 \text{ kJ/\$} \]

Conclusion & Comparison:

Least Expensive Source: Isooctane yields approximately \(7.4 \times 10^4 \text{ kJ/\$}\), while the cereal provides only \(4.1 \times 10^3 \text{ kJ/\$}\). Therefore, isooctane is by far the least expensive source of energy per dollar.

Nutritional Comparison: While isooctane produces nearly 18 times more thermodynamic energy per dollar through raw combustion, it holds zero nutritional value. Humans lack the metabolic pathways to break down hydrocarbons; isooctane is highly toxic if ingested. The breakfast cereal contains organic macromolecules (proteins, fats, and carbohydrates) that biological cells can safely metabolize via cellular respiration to generate biochemical ATP.

S7.4: Enthalpy

S38. Measured Heat ($q$) vs. Enthalpy Change ($\Delta H$)

The core difference lies in the scale of the quantities: \(q\) is an extensive path function representing the thermal energy exchanged in a specific trial run, whereas \(\Delta H\) is a normalized stoichiometric state property.

Measured Heat (\(q_{\text{rxn}}\)): In Example 5.5, \(q_{\text{rxn}} = -2.9 \times 10^3 \text{ J}\) (\(-2.9 \text{ kJ}\)). This value belongs explicitly to the scaling of that experimental trial, where only small fractions of a mole (\(0.0500 \text{ L} \times 1.00 \text{ M} = 0.0500 \text{ moles}\)) of \(\text{HCl}\) and \(\text{NaOH}\) were mixed.

Enthalpy Change (\(\Delta H\)): Enthalpy change represents the heat flow normalized per stoichiometric molar coefficient of the balanced equation. In this case, the equation specifies the reaction of exactly 1 full mole of \(\text{HCl}\) with 1 full mole of \(\text{NaOH}\).

To convert the trial's measured heat to the true thermodynamic \(\Delta H\), you must scale the energy up to a full molar foundation:

\[ \Delta H = \frac{q_{\text{rxn}}}{\text{moles of limiting reactant}} = \frac{-2.9 \text{ kJ}}{0.0500 \text{ mol}} = -58 \text{ kJ/mol} \]

S39. Calculating Enthalpy of Precipitation for \(\text{AgCl}\)

Step 1: Extract baseline parameters from Example 5.5's Check Your Learning module

Trial heat produced: \(q_{\text{produced}} = 1.34 \times 10^3 \text{ J} = 1.34 \text{ kJ}\)

Because the reaction is exothermic and transfers heat out to the surroundings, the trial reaction heat value is: \(q_{\text{rxn}} = -1.34 \text{ kJ}\)

Reactant volume: \(100 \text{ mL} = 0.100 \text{ L}\) of \(0.200 \text{ M } \text{AgNO}_3\)

Step 2: Calculate the moles of \(\text{AgNO}_3\) that reacted

\[ \text{Moles} = \text{Molarity (M)} \times \text{Volume (L)} \]

\[ \text{Moles of } \text{AgNO}_3 = 0.200 \text{ mol/L} \times 0.100 \text{ L} = 0.0200 \text{ mol} \]

Step 3: Normalize the enthalpy change per mole

\[ \Delta H = \frac{q_{\text{rxn}}}{\text{moles of } \text{AgNO}_3} \]

\[ \Delta H = \frac{-1.34 \text{ kJ}}{0.0200 \text{ mol}} = -67.0 \text{ kJ/mol} \]

The enthalpy of the reaction is \(-67.0 \text{ kJ/mol}\).

S40. Molar Enthalpy of Solution for \(\text{NH}_4\text{NO}_3\)

Step 1: Extract baseline parameters from Example 5.6

Measured heat of reaction: \(q_{\text{rxn}} = +1.0 \times 10^3 \text{ J} = +1.0 \text{ kJ}\) (endothermic)

Mass of solute dissolved: \(3.21 \text{ g of } \text{NH}_4\text{NO}_3\)

Step 2: Convert the solute mass to moles

The molar mass of Ammonium Nitrate (\(\text{NH}_4\text{NO}_3\)) is \(80.04 \text{ g/mol}\):

\[ \text{Moles} = \frac{3.21 \text{ g}}{80.04 \text{ g/mol}} = 0.0401 \text{ mol} \]

Step 3: Calculate the enthalpy change per mole

\[ \Delta H = \frac{q_{\text{rxn}}}{\text{moles}} \]

\[ \Delta H = \frac{+1.0 \text{ kJ}}{0.0401 \text{ mol}} = +25 \text{ kJ/mol} \]

The molar enthalpy of solution for \(\text{NH}_4\text{NO}_3\) is \(+25 \text{ kJ/mol}\).

S41. Enthalpy Calculation for the Potassium Chloride Dissolution

Step 1: Extract baseline parameters computed in Problem 23

Measured heat of reaction: \(q_{\text{rxn}} = +221.64 \text{ J} = +0.22164 \text{ kJ}\)

Mass of solute dissolved: \(0.500 \text{ g of } \text{KCl}\)

Step 2: Convert the solute mass to moles

The molar mass of Potassium Chloride (\(\text{KCl}\)) is \(74.55 \text{ g/mol}\):

\[ \text{Moles of } \text{KCl} = \frac{0.500 \text{ g}}{74.55 \text{ g/mol}} = 0.006707 \text{ mol} \]

Step 3: Calculate \(\Delta H\) per mole

\[ \Delta H = \frac{q_{\text{rxn}}}{\text{moles}} \]

\[ \Delta H = \frac{+0.22164 \text{ kJ}}{0.006707 \text{ mol}} = +33.0 \text{ kJ/mol} \]

The enthalpy of solution for \(\text{KCl}\) is \(+33.0 \text{ kJ/mol}\).

S42. Molar Enthalpy of Solution for \(\text{CaCl}_2\)

Step 1: Extract baseline parameters computed in Problem 24

Measured heat of reaction: \(q_{\text{rxn}} = -2174.3 \text{ J} = -2.1743 \text{ kJ}\)

Mass of solute dissolved: \(3.0 \text{ g of } \text{CaCl}_2\)

Step 2: Convert the solute mass to moles

The molar mass of Calcium Chloride (\(\text{CaCl}_2\)) is \(110.98 \text{ g/mol}\):

\[ \text{Moles of } \text{CaCl}_2 = \frac{3.0 \text{ g}}{110.98 \text{ g/mol}} = 0.02703 \text{ mol} \]

Step 3: Calculate \(\Delta H\) per mole

\[ \Delta H = \frac{q_{\text{rxn}}}{\text{moles}} \]

\[ \Delta H = \frac{-2.1743 \text{ kJ}}{0.02703 \text{ mol}} = -80.44 \text{ kJ/mol} \]

Rounding to two significant figures to lock step with the initial 3.0 g solute assignment:

The molar enthalpy of solution for \(\text{CaCl}_2\) is \(-80 \text{ kJ/mol}\).

S43. Real World Torch vs. Tabulated Enthalpy

The values listed in Table 5.2 are thermodynamic standard state quantities (\(\Delta H_c^\circ\)), which strictly require all reactants and products to be maintained at standard pressure (1 bar) and a reference temperature of exactly \(25^\circ\text{C}\).

In an operating oxyacetylene torch, two primary conditions cause the actual heat produced to deviate:

1. Physical States of Products: The standard thermodynamic value assumes that any water produced cools down and condenses entirely into liquid water (\(\text{H}_2\text{O}(l)\)), a process that releases its latent heat of vaporization. In an actual operating torch, the flame burns at temperatures exceeding \(3000^\circ\text{C}\), meaning the water leaves the reaction zone entirely as water vapor (\(\text{H}_2\text{O}(g)\)), bypassing that extra energy release.

2. Non-Standard Temperatures: The intense thermal energy of the flame environment keeps the local reactants and products far above the reference standard of \(25^\circ\text{C}\).

S44. Molar Scaling of Acetylene Combustion

From Table 5.2, the standard enthalpy of combustion for 1 mole of acetylene (\(\text{C}_2\text{H}_2\)) is \(-1301.1 \text{ kJ/mol}\).

\[ q = n \times \Delta H_c^\circ \]

\[ q = 4.00 \text{ mol} \times (-1301.1 \text{ kJ/mol}) = -5204.4 \text{ kJ} \]

The amount of heat produced is \(5.20 \times 10^3 \text{ kJ}\).

S45. Heat Produced by a Mass of Methanol

Step 1: Convert mass to moles

The molar mass of methanol (\(\text{CH}_3\text{OH}\)) is \(32.04 \text{ g/mol}\).

\[ n = \frac{125 \text{ g}}{32.04 \text{ g/mol}} = 3.9014 \text{ mol} \]

Step 2: Calculate heat using the table value (\(\Delta H_c^\circ = -726.1 \text{ kJ/mol}\))

\[ q = 3.9014 \text{ mol} \times (-726.1 \text{ kJ/mol}) = -2832.8 \text{ kJ} \]

Rounding to three significant figures:

The amount of heat produced is \(2.83 \times 10^3 \text{ kJ}\).

S46. Moles of Isooctane Required

From Table 5.2, the combustion of 1 mole of isooctane (\(\text{C}_8\text{H}_{18}\)) produces 5461 kJ of heat.

\[ n = \frac{\text{Target Heat}}{\Delta H_c^\circ} = \frac{-100 \text{ kJ}}{-5461 \text{ kJ/mol}} = 0.01831 \text{ mol} \]

The number of moles required is \(0.0183 \text{ mol}\).

S47. Mass of Carbon Monoxide Required

Step 1: Calculate moles of \(\text{CO}\) needed (\(\Delta H_c^\circ = -283.0 \text{ kJ/mol}\))

\[ n = \frac{-175 \text{ kJ}}{-283.0 \text{ kJ/mol}} = 0.61837 \text{ mol} \]

Step 2: Convert moles to mass

The molar mass of carbon monoxide (\(\text{CO}\)) is \(28.01 \text{ g/mol}\).

\[ m = 0.61837 \text{ mol} \times 28.01 \text{ g/mol} = 17.32 \text{ g} \]

Rounding to three significant figures:

The required mass of carbon monoxide is 17.3 g.

S48. Finding Enthalpy of Combustion from Experimental Data

Step 1: Convert the burning mass to moles

The molar mass of methane (\(\text{CH}_4\)) is \(16.04 \text{ g/mol}\).

\[ n = \frac{2.50 \text{ g}}{16.04 \text{ g/mol}} = 0.15586 \text{ mol} \]

Step 2: Calculate heat flow per mole

Since heat is produced (released), \(q_{\text{rxn}} = -125 \text{ kJ}\).

\[ \Delta H_c = \frac{q_{\text{rxn}}}{n} = \frac{-125 \text{ kJ}}{0.15586 \text{ mol}} = -801.99 \text{ kJ/mol} \]

Rounding to three significant figures:

The enthalpy of combustion under these conditions is \(-802 \text{ kJ/mol}\).

S49. Neutralization with Limiting Reactants and Temperature Shifts

Step 1: Identify the limiting reactant

\(\text{Moles of HCl} = 0.100 \text{ L} \times 0.250 \text{ M} = 0.0250 \text{ mol}\)

\(\text{Moles of NaOH} = 0.200 \text{ L} \times 0.150 \text{ M} = 0.0300 \text{ mol}\)

Since they react in a 1:1 stoichiometric ratio, \(\text{HCl}\) is the limiting reactant.

Step 2: Calculate the heat produced

The standard enthalpy of neutralization is \(-58 \text{ kJ/mol}\) (\(-58,000 \text{ J/mol}\)).

\[ q_{\text{rxn}} = n_{\text{limiting}} \times \Delta H_{\text{neut}} = 0.0250 \text{ mol} \times (-58,000 \text{ J/mol}) = -1450 \text{ J} \]

The heat produced by the mixture is 1450 J (or 1.45 kJ).

Step 3: Calculate the temperature increase

Total solution mass: \(m = (100 \text{ mL} \times 1.00 \text{ g/mL}) + (200 \text{ mL} \times 1.00 \text{ g/mL}) = 300 \text{ g}\)

Rearrange heat formula: \(\Delta T = \frac{q_{\text{absorbed}}}{m \cdot c_s}\)

\[ \Delta T = \frac{1450 \text{ J}}{300 \text{ g} \times 4.19 \text{ J/g}\cdot^\circ\text{C}} = \frac{1450}{1257} = 1.153^\circ\text{C} \]

The temperature will increase by \(1.15^\circ\text{C}\).

Assumptions: We assume that the calorimeter is perfectly insulated (no heat leaks to the room), and that the heat capacity and density of the final salt solution behave identically to those of pure water.

S50. Determining Calibration Constants for a Bomb Calorimeter

Step 1: Calculate the moles of carbon consumed

\[ n = \frac{0.562 \text{ g}}{12.01 \text{ g/mol}} = 0.04679 \text{ mol} \]

Step 2: Calculate the heat released by the sample (\(\Delta H_{c,\text{graphite}} = -393.5 \text{ kJ/mol}\))

\[ q_{\text{released}} = 0.04679 \text{ mol} \times 393.5 \text{ kJ/mol} = 18.412 \text{ kJ} = 18,412 \text{ J} \]

Step 3: Calculate the temperature change (\(\Delta T\))

\[ \Delta T = 27.93^\circ\text{C} - 26.74^\circ\text{C} = 1.19^\circ\text{C} \]

Step 4: Compute the total heat capacity (\(C_{\text{cal}}\))

\[ C_{\text{cal}} = \frac{q_{\text{absorbed}}}{\Delta T} = \frac{18.412 \text{ kJ}}{1.19^\circ\text{C}} = 15.47 \text{ kJ/}^\circ\text{C} \]

The heat capacity of the calorimeter and its contents is \(15.5 \text{ kJ/}^\circ\text{C}\) (or \(1.55 \times 10^4 \text{ J/}^\circ\text{C}\)).

S51. Comparing Mass Capacities of Refrigerant Liquids

Step 1: Standardize energy unit transformations

Convert the \(\text{SO}_2\) heat capacity from kilocalories to kilojoules (\(1 \text{ kcal} = 4.184 \text{ kJ}\)):

\[ \Delta H_{\text{vap},\text{SO}_2} = 6.00 \text{ kcal/mol} \times 4.184 \text{ kJ/kcal} = 25.104 \text{ kJ/mol} \]

Step 2: Calculate total heat removed by the baseline Freon load

Moles of \(\text{CCl}_2\text{F}_2\) (Molar mass = \(120.91 \text{ g/mol}\)):

\[ n = \frac{1000 \text{ g}}{120.91 \text{ g/mol}} = 8.2706 \text{ mol} \]

Total energy absorbed:

\[ q = 8.2706 \text{ mol} \times 17.4 \text{ kJ/mol} = 143.91 \text{ kJ} \]

Step 3: Solve for the equivalent mass of \(\text{SO}_2\)

Required moles of \(\text{SO}_2\):

\[ n_{\text{SO}_2} = \frac{143.91 \text{ kJ}}{25.104 \text{ kJ/mol}} = 5.7326 \text{ mol} \]

Convert to grams (Molar mass of \(\text{SO}_2 = 64.06 \text{ g/mol}\)):

\[ m = 5.7326 \text{ mol} \times 64.06 \text{ g/mol} = 367.23 \text{ g} \]

The required mass of sulfur dioxide is 367 g (or 0.367 kg).

S52. Heat Transfer: Water Cooling vs. Steam Cooling

Note: Use the standard specific heat values: \(c_{s,\text{steam}} = 1.864 \text{ J/g}\cdot^\circ\text{C}\) and \(c_{s,\text{liquid water}} = 4.184 \text{ J/g}\cdot^\circ\text{C}\)

Step 1: Calculate the heat released by the cooling steam (\(q_{\text{steam}}\))

The steam cools from \(110^\circ\text{C}\) to \(100^\circ\text{C}\), remaining a gas throughout this step (\(\Delta T = 100 - 110 = -10^\circ\text{C}\)).

\[ q_{\text{steam}} = m \cdot c_s \cdot \Delta T \]

\[ q_{\text{steam}} = 100 \text{ g} \times 1.864 \text{ J/g}\cdot^\circ\text{C} \times (-10.0^\circ\text{C}) = -1864 \text{ J} \]

Thus, the cooling steam provides exactly 1864 J of heat.

Step 2: Calculate the mass of liquid water required to release the same heat

We need the liquid water to release 1864 J (\(q_{\text{water}} = -1864 \text{ J}\)) when cooling from \(95.0^\circ\text{C}\) to \(35.0^\circ\text{C}\) (\(\Delta T = 35.0 - 95.0 = -60.0^\circ\text{C}\)).

\[ q_{\text{water}} = m \cdot c_s \cdot \Delta T \]

\[ -1864 \text{ J} = m \times 4.184 \text{ J/g}\cdot^\circ\text{C} \times (-60.0^\circ\text{C}) \]

\[ -1864 = -251.04 \cdot m \]

\[ m = \frac{-1864}{-251.04} = 7.425 \text{ g} \]

Rounding to three significant figures to match the temperature inputs:

A mass of 7.43 g of water will provide the same amount of heat.

S53. Enthalpies of Combustion vs. Enthalpies of Formation

By definition, a standard enthalpy of formation (\(\Delta H_f^\circ\)) is the enthalpy change for a reaction that creates exactly 1 mole of a pure substance from its constituent elements in their most stable, standard physical states.

Reviewing the chemical equations in Table 5.2, three combustion reactions perfectly match this criterion because their products are synthesized directly from pure elemental reactants:

1. Carbon Combustion:

\[ \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \]

This is the combustion of carbon, but it is also the standard formation reaction for **carbon dioxide gas (\(\text{CO}_2(g)\)).

2. Hydrogen Combustion:

\[ \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \]

This is the combustion of hydrogen gas, but it is also the standard formation reaction for liquid water (\(\text{H}_2\text{O}(l)\)).

3. Magnesium Combustion:

\[ \text{Mg}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{MgO}(s) \]

This is the combustion of magnesium metal, but it is also the standard formation reaction for solid magnesium oxide (\(\text{MgO}(s)\)).

4. Sulfur Combustion:

\[ \text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g) \]

This is the combustion of elemental sulfur, but it is also the standard formation reaction for sulfur dioxide gas (\(\text{SO}_2(g)\)).

Conclusion: The combustion enthalpies for carbon, hydrogen, magnesium, and sulfur are also standard enthalpies of formation.

Note: Reactions involving molecules like methane, methanol, or carbon monoxide do not qualify because those reactants are compounds, not basic standard-state elements

S54. Standard Enthalpy of Formation Alignment

Yes, they differ by a factor of 2.

Explanation: By definition, the standard enthalpy of formation (\(\Delta H_f^\circ\)) must represent the formation of exactly 1 mole of a compound from its constituent elements in their stable standard states: \[ \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(g) \quad \Delta H_f^\circ = -241.82 \text{ kJ/mol} \]

The chemical equation provided in the problem shows the formation of 2 moles of \(\text{H}_2\text{O}(g)\). Because enthalpy is an extensive property, its value scales directly with the molar quantities. Therefore, the \(\Delta H^\circ\) for the given reaction is twice the standard enthalpy of formation:

\[ \Delta H^\circ = 2 \times \Delta H_f^\circ = 2 \times (-241.82 \text{ kJ}) = -483.64 \text{ kJ} \]

S55.Decomposition of Mercury (II) Oxide

Step 1: Write the balanced chemical equation for the decomposition of 1 mole of \(\text{HgO}(s)\)

\[ \text{HgO}(s, \text{ red}) \rightarrow \text{Hg}(l) + \frac{1}{2}\text{O}_2(g) \]

Step 2: Retrieve standard formation values (\(\Delta H_f^\circ\)) from Appendix G

\(\Delta H_f^\circ \left[\text{HgO}(s, \text{ red})\right] = -90.83 \text{ kJ/mol}\)

\(\Delta H_f^\circ \left[\text{Hg}(l)\right] = 0 \text{ kJ/mol}\) (pure standard element)

\(\Delta H_f^\circ \left[\text{O}_2(g)\right] = 0 \text{ kJ/mol}\) (pure standard element)

Step 3: Apply Hess's Law

\[ \Delta H_{\text{rxn}}^\circ = \sum n \cdot \Delta H_f^\circ(\text{products}) - \sum m \cdot \Delta H_f^\circ(\text{reactants}) \]

\[ \Delta H_{\text{rxn}}^\circ = \left[ 1(0) + \frac{1}{2}(0) \right] - \left[ 1(-90.83 \text{ kJ}) \right] = +90.83 \text{ kJ} \]

Exactly \(90.83 \text{ kJ}\) of heat must be added to decompose one mole of red \(\text{HgO}(s)\).

S56. Combustion of Manganese

Step 1: Write the chemical equation for the combustion of exactly 1 mole of \(\text{Mn}(s)\)

\[ \text{Mn}(s) + \frac{2}{3}\text{O}_2(g) \rightarrow \frac{1}{3}\text{Mn}_3\text{O}_4(s) \]

Step 2: Retrieve the standard enthalpy of formation from Appendix G

\(\Delta H_f^\circ \left[\text{Mn}_3\text{O}_4(s)\right] = -1378.8 \text{ kJ/mol}\)

Step 3: Compute the heat change relative to 1 mole of \(\text{Mn}\)

Because the balanced equation produces \(\frac{1}{3}\) of a mole of \(\text{Mn}_3\text{O}_4\), we scale the molar formation value accordingly:

\[ \Delta H_{\text{rxn}}^\circ = \frac{1}{3} \times (-1378.8 \text{ kJ}) = -459.6 \text{ kJ} \]

Exactly \(459.6 \text{ kJ}\) of heat will be released.

S57. Combustion of Iron

Step 1: Write the chemical equation for the combustion of exactly 1 mole of \(\text{Fe}(s)\)

\[ \text{Fe}(s) + \frac{3}{4}\text{O}_2(g) \rightarrow \frac{1}{2}\text{Fe}_2\text{O}_3(s) \]

Step 2: Retrieve the standard enthalpy of formation from Appendix G

\(\Delta H_f^\circ \left[\text{Fe}_2\text{O}_3(s)\right] = -824.2 \text{ kJ/mol}\)

Step 3: Compute the heat change

\[ \Delta H_{\text{rxn}}^\circ = \frac{1}{2} \times (-824.2 \text{ kJ}) = -412.1 \text{ kJ} \]

Exactly \(412.1 \text{ kJ}\) of heat will be released.

S58. Multi-Step Production Enthalpy of Nitric Acid

To find the total energy change to produce exactly 1 mole of \(\text{HNO}_3(aq)\), we normalize each structural reaction step to establish a coherent reaction net path via Hess's Law.

Step 3 establishes our target asset: It yields \(2 \text{ moles of } \text{HNO}_3\). To scale this step down to 1 mole, we multiply the entire step by \(\frac{1}{2}\): \[ \frac{3}{2}\text{NO}_2(g) + \frac{1}{2}\text{H}_2\text{O}(l) \rightarrow \mathbf{1}\text{HNO}_3(aq) + \frac{1}{2}\text{NO}(g) \quad \Delta H_3 = \frac{1}{2}(-139 \text{ kJ}) = -69.5 \text{ kJ} \]

Step 2 supplies the \(\text{NO}_2\) intermediate: We need \(\frac{3}{2}\text{NO}_2\) to feed the reaction above. Since Step 2 natively produces \(2\text{NO}_2\), we multiply it by \(\frac{3}{4}\): \[ \frac{3}{2}\text{NO}(g) + \frac{3}{4}\text{O}_2(g) \rightarrow \frac{3}{2}\text{NO}_2(g) \quad \Delta H_2 = \frac{3}{4}(-113 \text{ kJ}) = -84.75 \text{ kJ} \]

Step 1 supplies the \(\text{NO}\) intermediate: To provide the \(\frac{3}{2}\text{NO}\) consumed in the step above (while subtracting the \(\frac{1}{2}\text{NO}\) recycled out as a product in Step 3), the net \(\text{NO}\) required from Step 1 is exactly \(1\text{ mole of } \text{NO}\). Since Step 1 natively produces \(4\text{NO}\), we multiply it by \(\frac{1}{4}\):

\[ 1\text{NH}_3(g) + \frac{5}{4}\text{O}_2(g) \rightarrow 1\text{NO}(g) + \frac{6}{4}\text{H}_2\text{O}(l) \quad \Delta H_1 = \frac{1}{4}(-907 \text{ kJ}) = -226.75 \text{ kJ} \]

Sum the individual scaled enthalpy components:

\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 + \Delta H_3 \]

\[ \Delta H_{\text{total}} = -226.75 \text{ kJ} + (-84.75 \text{ kJ}) + (-69.5 \text{ kJ}) = -381 \text{ kJ} \]

The total energy change to produce 1 mole of aqueous nitric acid is \(-381 \text{ kJ}\).

S59. Combustion Comparison: Graphite vs. Diamond

The combustion of diamond produces more heat.

Explanation:

The transition reaction shows that converting graphite to diamond is an endothermic process (\(\Delta H^\circ > 0\)). This means that diamond possesses higher internal chemical potential energy than graphite by exactly \(1.90 \text{ kJ/mol}\).

When both substances are burned completely to form the same low-energy product (\(\text{CO}_2(g)\)), diamond begins the path from a higher potential energy starting line. Consequently, it drops further down the energy scale, releasing that extra internal potential energy as supplemental thermal output. The combustion of diamond will release exactly \(1.90 \text{ kJ}\) more heat per mole than the combustion of graphite.

S60. Enthalpy of Vaporization for Water via Appendix G

Step 1: Retrieve standard values from Appendix G

\(\Delta H_f^\circ \left[\text{H}_2\text{O}(l)\right] = -285.83 \text{ kJ/mol}\)

\(\Delta H_f^\circ \left[\text{H}_2\text{O}(g)\right] = -241.82 \text{ kJ/mol}\)

Step 2: Apply Hess's Law

\[ \Delta H_{\text{vap}}^\circ = \Delta H_f^\circ\left[\text{H}_2\text{O}(g)\right] - \Delta H_f^\circ\left[\text{H}_2\text{O}(l)\right] \]

\[ \Delta H_{\text{vap}}^\circ = -241.82 \text{ kJ/mol} - (-285.83 \text{ kJ/mol}) = +44.01 \text{ kJ/mol} \]

Exactly \(44.01 \text{ kJ}\) of heat is required to evaporate one mole of water under standard state conditions.

S61. Comparing Heat Production Across Phase States

The reaction that produces solid osmium tetroxide, \(\text{OsO}_4(s)\), releases more heat.

Explanation: Both reactions start with identical reactants in the same states, but one ends with a solid product and the other ends with a gas product. Converting a solid to a gas (sublimation) requires an investment of energy (\(\Delta H = +56.4 \text{ kJ}\)).

Therefore, when the system falls from its high-energy elemental reactants down to the products, the pathway forming the gas must withhold \(56.4 \text{ kJ/mol}\) of energy to keep the molecules in the gaseous state. The path forming the lower-energy solid drops further down the energy scale, releasing that extra energy as additional heat output to the surroundings.

S62. Hess's Law Application: Antimony Pentachloride Synthesis

We can find the target enthalpy change by adding the two given equations directly together.

The intermediate \(\text{SbCl}_3(g)\) cancels out completely on both sides, yielding the target equation: \[ \text{Sb}(s) + \frac{5}{2}\text{Cl}_2(g) \rightarrow \text{SbCl}_5(g) \quad \Delta H^\circ_{298} = -394 \text{ kJ} \]

S63. Hess's Law Application: Zinc Sulfate Formation

The solid intermediate \(\text{ZnS}(s)\) cancels out from the products of the first step and the reactants of the second step: \[ \text{Zn}(s) + \text{S}(s) + 2\text{O}_2(g) \rightarrow \text{ZnSO}_4(s) \quad \Delta H^\circ_{298} = -982.8 \text{ kJ} \]

S64. Hess's Law Application: Calomel Decomposition

We need \(\text{Hg}_2\text{Cl}_2(s)\) as a reactant. Reverse Equation 2 (change the sign of \(\Delta H\)):

\[ \text{Hg}_2\text{Cl}_2(s) \rightarrow \text{Hg}(l) + \text{HgCl}_2(s) \quad \Delta H = +41.2 \text{ kJ} \]

We need to eliminate the intermediate compound \(\text{HgCl}_2(s)\) from our final net reaction. Reverse Equation 1 to place it as a reactant so it cancels out:

\[ \text{HgCl}_2(s) \rightarrow \text{Hg}(l) + \text{Cl}_2(g) \quad \Delta H = +224 \text{ kJ} \]

The intermediate solid mercury(II) chloride cancels out cleanly: \[ \text{Hg}_2\text{Cl}_2(s) \rightarrow 2\text{Hg}(l) + \text{Cl}_2(g) \quad \Delta H = +265.2 \text{ kJ} \]

S65. Hess's Law Application: Cobalt Oxide Decomposition

Notice that the target reaction only involves \(\text{Co}_3\text{O}_4(s)\), \(\text{Co}(s)\), and \(\text{O}_2(g)\). The first equation involving \(\text{CoO}(s)\) is extra information and is not needed.

To get the target equation, simply reverse **Equation 2** (which changes the mathematical sign of its \(\Delta H^\circ\)):

\[ \text{Co}_3\text{O}_4(s) \rightarrow 3\text{Co}(s) + \text{O}_2(g) \quad \Delta H^\circ_{298} = +177.5 \text{ kJ} \]

S66. Hess's Law Application: Standard Enthalpy of Formation for \(\text{NO}(g)\)

Step 1: Set up the target chemical equation

By definition, the formation reaction for 1 mole of \(\text{NO}(g)\) from its pure elemental states is:

\[ \frac{1}{2}\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{NO}(g) \]

Step 2: Manipulate the given equations

Keep Equation 1 as is to get \(\text{N}_2(g)\) on the reactant side:

\[ \text{N}_2(g) + 2\text{O}_2(g) \rightarrow 2\text{NO}_2(g) \quad \Delta H^\circ = +66.4 \text{ kJ}\]

Reverse Equation 2 to place \(\text{NO}(g)\) on the product side:

\[ 2\text{NO}_2(g) \rightarrow 2\text{NO}(g) + \text{O}_2(g) \quad \Delta H^\circ = +114.1 \text{ kJ}\]

Because this represents the formation of 2 moles of \(\text{NO}\), divide the entire result by 2 to obtain the standard molar enthalpy of formation:

\[ \Delta H_f^\circ = \frac{180.5 \text{ kJ}}{2} = +90.25 \text{ kJ/mol} \]

S67. Standard Enthalpy Changes via Hess's Law

a) \(\Delta H^\circ = [2(90.25)] - [0 + 0] =\) \(+180.5 \text{ kJ}\)

b) \(\Delta H^\circ = [-662.75] - [0 + 0] =\) \(-662.8 \text{ kJ}\)

c) \(\Delta H^\circ = [2(0) + 3(-285.83)] - [-824.2 + 3(0)] = -857.49 + 824.2 =\) \(-33.3 \text{ kJ}\)

d) \(\Delta H^\circ = [(-1216.03) + (-241.82)] - [2(-484.93) + (-393.51)] = -1457.85 - (-1363.37) =\) \(-94.5 \text{ kJ}\)

S68. Standard Enthalpy Changes via Hess's Law

a) \(\Delta H^\circ = [-1615.0] - [0 + 0] =\) \(-1615.0 \text{ kJ}\)

b) \(\Delta H^\circ = [-484.3] - [0 + 0 + 0] =\) \(-484.3 \text{ kJ}\)

c) \(\Delta H^\circ = [(135.5) + (-45.9)] - [-74.6 + 0] = 89.6 - (-74.6) =\) \(+164.2 \text{ kJ}\)

d) \(\Delta H^\circ = [(-95.7) + (-58.2)] - [+116.7 + 3(0)] = -153.9 - 116.7 =\) \(-270.6 \text{ kJ}\)

S69. Enthalpy Changes for Metal Preparation Reactions

a) \(\Delta H^\circ = [4(0) + 0] - [2(-31.05)] = 0 - (-62.1) =\) \(+62.1 \text{ kJ}\)

b) \(\Delta H^\circ = [0 + (-393.51)] - [(-285.8) + (-110.53)] = -393.51 - (-396.33) =\) \(+2.8 \text{ kJ}\)

c) \(\Delta H^\circ = [2(0) + 3(-285.83)] - [-1139.7 + 3(0)] = -857.49 - (-1139.7) =\) \(+282.2 \text{ kJ}\)

d) \(\Delta H^\circ = [-1675.7 + 2(0)] - [2(0) + (-824.2)] = -1675.7 - (-824.2) =\) \(-851.5 \text{ kJ}\)

S70. Decomposition of Hydrogen Peroxide

Step 1: Retrieve standard values from Appendix G

\(\Delta H_f^\circ\ [\text{H}_2\text{O}_2(l)] = -187.78 \text{ kJ/mol}\)

\(\Delta H_f^\circ\ [\text{H}_2\text{O}(g)] = -241.82 \text{ kJ/mol}\)

\(\Delta H_f^\circ\ [\text{O}_2(g)] = 0 \text{ kJ/mol}\)

Step 2: Apply Hess's Law for the given balanced equation (2 moles of \(\text{H}_2\text{O}_2\))

\[ \Delta H_{\text{rxn}}^\circ = [2(-241.82) + 0] - [2(-187.78)] = -483.64 - (-375.56) = -108.08 \text{ kJ} \]

Step 3: Scale down to exactly 1 mole of \(\text{H}_2\text{O}_2\)

\[ \text{Heat produced per mole} = \frac{-108.08 \text{ kJ}}{2 \text{ mol}} = -54.04 \text{ kJ/mol} \]

Exactly \(54.04 \text{ kJ}\) of heat is produced per mole of decomposed \(\text{H}_2\text{O}_2(l)\)

S71. Enthalpy of Combustion of Propane

Step 1: Write the balanced combustion equation

\[ \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \]

Step 2: Retrieve remaining standard values from Appendix G

\(\Delta H_f^\circ\ [\text{CO}_2(g)] = -393.51 \text{ kJ/mol}\)

\(\Delta H_f^\circ\ [\text{H}_2\text{O}(g)] = -241.82 \text{ kJ/mol}\)

Step 3: Calculate \(\Delta H_{\text{comb}}^\circ\)

\[ \Delta H_{\text{comb}}^\circ = [3(-393.51) + 4(-241.82)] - [1(-104) + 5(0)] \]

\[ \Delta H_{\text{comb}}^\circ = [-1180.53 - 967.28] - [-104] = -2147.81 + 104 = -2043.81 \text{ kJ/mol} \]

Rounding to match the integer precision of the given propane data:

The enthalpy of combustion of propane is \(-2044 \text{ kJ/mol}\).

S72. Enthalpy of Combustion of Butane

Step 1: Write the balanced combustion equation

\[ \text{C}_4\text{H}_{10}(g) + \frac{13}{2}\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 5\text{H}_2\text{O}(g) \]

Step 2: Calculate \(\Delta H_{\text{comb}}^\circ\)

\[ \Delta H_{\text{comb}}^\circ = [4(-393.51) + 5(-241.82)] - [1(-126) + 0] \]

\[ \Delta H_{\text{comb}}^\circ = [-1574.04 - 1209.1] + 126 = -2783.14 + 126 = -2657.14 \text{ kJ/mol} \]

Rounding to the units place: The enthalpy of combustion of butane is \(-2657 \text{ kJ/mol}\).

S73. Fuel Efficiency: Propane vs. Butane per Gram

For Propane (\(\text{C}_3\text{H}_8\)): Molar mass = \(44.11 \text{ g/mol}\); \(\Delta H_{\text{comb}}^\circ = -2044 \text{ kJ/mol}\)

\[ \text{Heat per gram} = \frac{2044 \text{ kJ}}{44.11 \text{ g}} = 46.34 \text{ kJ/g} \]

For Butane (\(\text{C}_4\text{H}_{10}\)): Molar mass = \(58.14 \text{ g/mol}\); \(\Delta H_{\text{comb}}^\circ = -2657 \text{ kJ/mol}\)

\[ \text{Heat per gram} = \frac{2657 \text{ kJ}}{58.14 \text{ g}} = 45.70 \text{ kJ/g} \]

Conclusion: Propane produces slightly more heat per gram (\(46.3 \text{ kJ/g}\)) compared to butane (\(45.7 \text{ kJ/g}\)).

S74. Production of Titanium Dioxide

Step 1: Retrieve standard values from Appendix G

\(\Delta H_f^\circ\ [\text{TiCl}_4(g)] = -763.2 \text{ kJ/mol}\)

\(\Delta H_f^\circ\ [\text{H}_2\text{O}(g)] = -241.82 \text{ kJ/mol}\)

\(\Delta H_f^\circ\ [\text{TiO}_2(s)] = -944.0 \text{ kJ/mol}\)

\(\Delta H_f^\circ\ [\text{HCl}(g)] = -92.31 \text{ kJ/mol}\)

Step 2: Calculate \(\Delta H_{\text{rxn}}^\circ\)

\[ \Delta H_{\text{rxn}}^\circ = [1(-944.0) + 4(-92.31)] - [1(-763.2) + 2(-241.82)] \]

\[ \Delta H_{\text{rxn}}^\circ = [-944.0 - 369.24] - [-763.2 - 483.64] \]

\[ \Delta H_{\text{rxn}}^\circ = -1313.24 - (-1246.84) = -66.4 \text{ kJ} \]

Exactly \(66.4 \text{ kJ}\) of heat is evolved.

S75. Water Gas and Methanol Synthesis Roadmap

\(\Delta H_f^\circ\ [\text{CO}(g)] = -110.53 \text{ kJ/mol}\); \(\Delta H_f^\circ\ [\text{CH}_3\text{OH}(g)] = -201.0 \text{ kJ/mol}\); \(\Delta H_f^\circ\ [\text{CH}_3\text{OH}(l)] = -239.2 \text{ kJ/mol}\)

a) Water Gas Generation Enthalpy:

\[ \Delta H^\circ = [1(-110.53) + 0] - [0 + 1(-241.82)] = -110.53 + 241.82 = \mathbf{+131.29 \text{ kJ}} \]

b) Gaseous Methanol Formation and Condensation:

Synthesis \(\Delta H^\circ\): \([-201.0] - [2(0) + (-110.53)] = -201.0 + 110.53 = \mathbf{-90.5 \text{ kJ}}\)

Condensation (\(\text{CH}_3\text{OH}(g) \rightarrow \text{CH}_3\text{OH}(l)\)):** \([-239.2] - [-201.0] = \mathbf{-38.2 \text{ kJ}}\)

c) Combustion of Liquid Methanol to Gaseous Water:

Equation: \(\text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)\)

\[ \Delta H^\circ = [1(-393.51) + 2(-241.82)] - [1(-239.2) + 0] \]

\[ \Delta H^\circ = [-393.51 - 483.64] + 239.2 = -877.15 + 239.2 = \mathbf{-638.0 \text{ kJ}} \]

S76. Acetylene Miner's Lamp Reaction

Step 1: Standardize energy unit transformations

\[ \Delta H_f^\circ\ [\text{CaC}_2(s)] = -15.14 \text{ kcal/mol} \times 4.184 \text{ kJ/kcal} = -63.35 \text{ kJ/mol} \]

Step 2: Collect data parameters

\(\Delta H_f^\circ\ [\text{H}_2\text{O}(l)] = -285.83 \text{ kJ/mol}\); \(\Delta H_f^\circ\ [\text{Ca(OH)}_2(s)] = -985.2 \text{ kJ/mol}\); \(\Delta H_f^\circ\ [\text{C}_2\text{H}_2(g)] = +227.4 \text{ kJ/mol}\)

Step 3: Calculate \(\Delta H_{\text{rxn}}^\circ\)

\[ \Delta H_{\text{rxn}}^\circ = [1(-985.2) + 1(227.4)] - [1(-63.35) + 2(-285.83)] \]

\[ \Delta H_{\text{rxn}}^\circ = [-757.8] - [-63.35 - 571.66] = -757.8 - (-635.01) = -122.8 \text{ kJ} \]

The standard enthalpy of the reaction is \(-122.8 \text{ kJ}\).

S77. Mass Efficiency Comparison of Gaseous Fuels

\(\text{CO}(g)\): Molar mass = \(28.01 \text{ g/mol}\); \(\Delta H_c^\circ = -283.0 \text{ kJ/mol}\)

\[ \text{Heat per gram} = \frac{283.0 \text{ kJ}}{28.01 \text{ g}} = 10.10 \text{ kJ/g} \]

\(\text{CH}_4(g)\): Molar mass = \(16.04 \text{ g/mol}\); \(\Delta H_c^\circ = -890.8 \text{ kJ/mol}\)

\[ \text{Heat per gram} = \frac{890.8 \text{ kJ}}{16.04 \text{ g}} = 55.54 \text{ kJ/g} \]

\(\text{C}_2\text{H}_2(g)\): Molar mass = \(26.04 \text{ g/mol}\); \(\Delta H_c^\circ = -1301.1 \text{ kJ/mol}\)

\[ \text{Heat per gram} = \frac{1301.1 \text{ kJ}}{26.04 \text{ g}} = 49.97 \text{ kJ/g} \]

Conclusion:

Methane (\(\text{CH}_4\)) produces the greatest amount of heat per gram (\(55.5 \text{ kJ/g}\)).

S78. Fuel Replacement: Hard Coal vs. Gasoline Equivalent

Step 1: State target heat output value

The total heat needed from the coal is equal to the heat of 1.0 gallon of gasoline:

\[ q = 1.28 \times 10^5 \text{ kJ} \]

Step 2: Calculate the mass of coal required in grams

\[ \text{Mass of coal} = \frac{1.28 \times 10^5 \text{ kJ}}{35 \text{ kJ/g}} = 3657.14 \text{ g} \]

Step 3: Convert the mass into kilograms

\[ \text{Mass in kg} = \frac{3657.14 \text{ g}}{1000 \text{ g/kg}} = 3.657 \text{ kg} \implies 3.7 \text{ kg} \]

Exactly 3.7 kg of hard coal is required.

Note: The density metrics provided in the problem statement are extra background details and are not needed because the total heat output per gallon of gasoline was already explicitly stated

S79. Ethanol and Gasoline Mileage Thermodynamics

a) Molar Enthalpy of Combustion:

Balanced Equation:

\[ \text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(g) \]

Appendix G Data: \(\Delta H_f^\circ \left[\text{C}_2\text{H}_5\text{OH}(l)\right] = -277.6 \text{ kJ/mol}\); \(\Delta H_f^\circ \left[\text{CO}_2(g)\right] = -393.51 \text{ kJ/mol}\); \(\Delta H_f^\circ \left[\text{H}_2\text{O}(g)\right] = -241.82 \text{ kJ/mol}\)

Calculate \(\Delta H_{\text{comb}}^\circ\):

\[ \Delta H_{\text{comb}}^\circ = [2(-393.51) + 3(-241.82)] - [-277.6 + 0] \]

\[ \Delta H_{\text{comb}}^\circ = [-787.02 - 725.46] + 277.6 = -1512.48 + 277.6 = \mathbf{-1234.9 \text{ kJ/mol}} \]

b) Enthalpy of Combustion for 1 L of Ethanol:

Mass of 1 L (1000 mL) of ethanol: \(m = 1000 \text{ mL} \times 0.7893 \text{ g/mL} = 789.3 \text{ g}\)

Moles of ethanol (Molar mass = \(46.07 \text{ g/mol}\)): \(n = \frac{789.3 \text{ g}}{46.07 \text{ g/mol}} = 17.133 \text{ mol}\)

Heat released:

\[ q = 17.133 \text{ mol} \times (-1234.9 \text{ kJ/mol}) = \mathbf{-2.116 \times 10^4 \text{ kJ}} \]

c) Mileage Comparison (Ethanol vs. Gasoline/Octane):

Step 1: Calculate the combustion heat of 1 L of n-octane

Balanced Equation: \(\text{C}_8\text{H}_{18}(l) + \frac{25}{2}\text{O}_2(g) \rightarrow 8\text{CO}_2(g) + 9\text{H}_2\text{O}(g)\)

Appendix G Data: \(\Delta H_f^\circ \left[\text{C}_8\text{H}_{18}(l)\right] = -208.4 \text{ kJ/mol}\)

Molar \(\Delta H_{\text{comb}}^\circ = [8(-393.51) + 9(-241.82)] - [-208.4] = -5324.46 + 208.4 = -5116.1 \text{ kJ/mol}\)

Mass of 1 L of octane: \(1000 \text{ mL} \times 0.7025 \text{ g/mL} = 702.5 \text{ g}\)

Moles of octane (Molar mass = \(114.23 \text{ g/mol}\)): \(n = \frac{702.5 \text{ g}}{114.23 \text{ g/mol}} = 6.150 \text{ mol}\)

Total heat from 1 L of gasoline: \(q = 6.150 \text{ mol} \times (-5116.1 \text{ kJ/mol}) = -3.146 \times 10^4 \text{ kJ}\)

Step 2: Compare the energy capacities

\[ \text{Ratio} = \frac{3.146 \times 10^4 \text{ kJ (Gasoline)}}{2.116 \times 10^4 \text{ kJ (Ethanol)}} = 1.487 \]

An automobile could travel 1.49 times farther (or approximately \(49\%\) farther) on 1 L of gasoline than on 1 L of ethanol.

S80. Rocket Fuel Performance Analysis

1. Diborane (\(\text{B}_2\text{H}_6\)): \(\text{B}_2\text{H}_6(g) + 3\text{O}_2(g) \rightarrow \text{B}_2\text{O}_3(s) + 3\text{H}_2\text{O}(g)\)

Appendix G Data: \(\Delta H_f^\circ \left[\text{B}_2\text{H}_6(g)\right] = +36.4 \text{ kJ/mol}\); \(\Delta H_f^\circ \left[\text{B}_2\text{O}_3(s)\right] = -1273.5 \text{ kJ/mol}\); \(\Delta H_f^\circ \left[\text{H}_2\text{O}(g)\right] = -241.82 \text{ kJ/mol}\)

Molar \(\Delta H_{\text{comb}}^\circ = [-1273.5 + 3(-241.82)] - [+36.4] = -2035.46 \text{ kJ/mol}\)

Mass efficiency (Molar mass = \(27.67 \text{ g/mol}\)): \(\frac{2035.46 \text{ kJ}}{27.67 \text{ g}} = \mathbf{73.56 \text{ kJ/g}}\)

2. Methane (\(\text{CH}_4\)): \(\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)\)

Molar \(\Delta H_{\text{comb}}^\circ = [-393.51 + 2(-241.82)] - [-74.6] = -802.55 \text{ kJ/mol}\)

Mass efficiency (Molar mass = \(16.04 \text{ g/mol}\)): \(\frac{802.55 \text{ kJ}}{16.04 \text{ g}} = \mathbf{50.03 \text{ kJ/g}}\)

3. Hydrazine (\(\text{N}_2\text{H}_4\)): \(\text{N}_2\text{H}_4(l) + \text{O}_2(g) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(g)\)

Appendix G Data: \(\Delta H_f^\circ \left[\text{N}_2\text{H}_4(l)\right] = +50.63 \text{ kJ/mol}\)

Molar \(\Delta H_{\text{comb}}^\circ = [0 + 2(-241.82)] - [+50.63] = -534.27 \text{ kJ/mol}\)

Mass efficiency (Molar mass = \(32.05 \text{ g/mol}\)): \(\frac{534.27 \text{ kJ}}{32.05 \text{ g}} = \mathbf{16.67 \text{ kJ/g}}\)

Conclusion:

Diborane (\(\text{B}_2\text{H}_6\)) offers the best possibility as a rocket fuel because it releases the greatest amount of heat per gram (\(73.6 \text{ kJ/g}\)).

S81. Chromium Combustion and Unlabeled Synthesis Enthalpies

Solution 80a:

Balanced Equation: \(4\text{Cr}(s) + 3\text{O}_2(g) \rightarrow 2\text{Cr}_2\text{O}_3(s)\)

Appendix G Data: \(\Delta H_f^\circ \left[\text{Cr}_2\text{O}_3(s)\right] = -1139.7 \text{ kJ/mol}\)

Moles of \(\text{Cr}\) (Molar mass = \(52.00 \text{ g/mol}\)): \(n = \frac{1.25 \text{ g}}{52.00 \text{ g/mol}} = 0.02404 \text{ mol}\)

Total heat produced:

\[ q = 0.02404 \text{ mol Cr} \times \left( \frac{-1139.7 \text{ kJ} \times 2}{4 \text{ mol Cr}} \right) = \mathbf{-13.7 \text{ kJ}} \]

Solution 80b:

Appendix G Data: \(\Delta H_f^\circ \left[\text{C}_2\text{H}_4(g)\right] = +52.4 \text{ kJ/mol}\); \(\Delta H_f^\circ \left[\text{H}_2\text{O}(g)\right] = -241.82 \text{ kJ/mol}\); \(\Delta H_f^\circ \left[\text{C}_2\text{H}_5\text{OH}(l)\right] = -277.6 \text{ kJ/mol}\)

Calculate \(\Delta H^\circ\):

\[ \Delta H^\circ = [-277.6] - [52.4 + (-241.82)] = -277.6 - (-189.42) = \mathbf{-88.2 \text{ kJ}} \]

S82. Biological Glucose Metabolism

The molar mass of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) is \(180.16 \text{ g/mol}\).

\[ \text{Moles in 1.0 g} = \frac{1.0 \text{ g}}{180.16 \text{ g/mol}} = 0.005551 \text{ mol} \]

a) Heat in Kilojoules:

\[ q = 0.005551 \text{ mol} \times (-2816 \text{ kJ/mol}) = \mathbf{-15.6 \text{ kJ}} \]

b) Heat in Nutritional Calories (kcal):

\[ \text{Calories} = \frac{15.63 \text{ kJ}}{4.184 \text{ kJ/kcal}} = \mathbf{3.7 \text{ Calories}} \]

S83. Complete Propane Combustion Analysis

a) Balanced Equation:

\[ \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \]

b) Volume of Air Required:

Moles of propane (Molar mass = \(44.1 \text{ g/mol}\)): \(n = \frac{25.0 \text{ g}}{44.1 \text{ g/mol}} = 0.5669 \text{ mol}\)

Moles of \(\text{O}_2\) needed: \(0.5669 \text{ mol} \times 5 = 2.8345 \text{ mol}\)

Mass of \(\text{O}_2\) needed: \(2.8345 \text{ mol} \times 32.00 \text{ g/mol} = 90.704 \text{ g}\)

Volume of air needed based on the density hint:

\[ \text{Volume of air} = \frac{90.704 \text{ g }\text{O}_2}{0.275 \text{ g }\text{O}_2/\text{L air}} = \mathbf{330. \text{ L air}} \]

c) Enthalpy of Formation (\(\Delta H_f^\circ\)) for Propane:

\[ \Delta H_{\text{comb}}^\circ = [3\Delta H_f^\circ(\text{CO}_2) + 4\Delta H_f^\circ(\text{H}_2\text{O})] - \Delta H_f^\circ(\text{C}_3\text{H}_8) \]

\[ -2219.2 = [3(-393.5) + 4(-285.8)] - \Delta H_f^\circ(\text{C}_3\text{H}_8) \] \[ -2219.2 = [-1180.5 - 1143.2] - \Delta H_f^\circ(\text{C}_3\text{H}_8) \]

\[ -2219.2 = -2323.7 - \Delta H_f^\circ(\text{C}_3\text{H}_8) \]

\[ \Delta H_f^\circ(\text{C}_3\text{H}_8) = -2323.7 + 2219.2 = \mathbf{-104.5 \text{ kJ/mol}} \]

d) Water Temperature Increase:

Heat released by 25.0 g (\(0.5669 \text{ mol}\)) of propane:

\[ q = 0.5669 \text{ mol} \times 2219.2 \text{ kJ/mol} = 1258.1 \text{ kJ} = 1,258,100 \text{ J} \]

Calculate \(\Delta T\) for \(4.00 \text{ kg} = 4000 \text{ g}\) of water:

\[ \Delta T = \frac{q}{m \cdot c_s} = \frac{1,258,100 \text{ J}}{4000 \text{ g} \times 4.184 \text{ J/g}\cdot^\circ\text{C}} = \frac{1,258,100}{16,736} = \mathbf{75.2^\circ\text{C}} \]

S84. Comprehensive Home Heating Energy Analysis

First, calculate the total gross combustion energy ($q_{\text{total}}$) required from the fuel before considering the furnace's thermal tax.

Net heat needed = \(3500\text{ kWh}\)

Efficiency = \(89\% = 0.89\)

\[ q_{\text{total}} = \frac{3500\text{ kWh}}{0.89} = 3932.58\text{ kWh} \]

\[ q_{\text{total}} = 3932.58\text{ kWh} \times (3.6 \times 10^6\text{ J/kWh}) = 1.4157 \times 10^{10}\text{ J} = 1.4157 \times 10^7\text{ kJ} \]

a) Volume of Methane Required (\(\text{ft}^3\)):

Step 1: The standard molar enthalpy of combustion for methane (\(\text{CH}_4\)) is \(-890.8\text{ kJ/mol}\). Find the moles of methane burned:

\[ n = \frac{1.4157 \times 10^7\text{ kJ}}{890.8\text{ kJ/mol}} = 15,893.15\text{ mol} \]

Step 2: Convert moles to mass (Molar mass of \(\text{CH}_4 = 16.04\text{ g/mol}\)):

\[ m = 15,893.15\text{ mol} \times 16.04\text{ g/mol} = 254,926\text{ g} \]

Step 3: Calculate volume in Liters using the density (\(0.681\text{ g/L}\)):

\[ V = \frac{254,926\text{ g}}{0.681\text{ g/L}} = 374,341\text{ L} \]

Step 4: Convert Liters to cubic feet (\(1\text{ ft}^3 = 28.3168\text{ L}\)):

\[ V = \frac{374,341\text{ L}}{28.3168\text{ L/ft}^3} = 13,220\text{ ft}^3 \implies \mathbf{1.32 \times 10^4\text{ ft}^3} \]

b) Volume of Liquid Propane Replacing Methane (\(\text{gal}\)):

Step 1: Find moles of propane needed (\(\Delta H_c^\circ = -2219\text{ kJ/mol}\)):

\[ n = \frac{1.4157 \times 10^7\text{ kJ}}{2219\text{ kJ/mol}} = 6379.9\text{ mol} \]

Step 2: Convert to mass (Molar mass of \(\text{C}_3\text{H}_8 = 44.11\text{ g/mol}\)):

\[ m = 6379.9\text{ mol} \times 44.11\text{ g/mol} = 281,417\text{ g} \]

Step 3: Convert to volume in milliliters using liquid density (\(0.5318\text{ g/mL}\)):

\[ V = \frac{281,417\text{ g}}{0.5318\text{ g/mL}} = 529,178\text{ mL} = 529.18\text{ L} \]

Step 4: Convert Liters to gallons (\(1\text{ gal} = 3.78541\text{ L}\)):

\[ V = \frac{529.18\text{ L}}{3.78541\text{ L/gal}} = 139.79\text{ gal} \implies \mathbf{1.40 \times 10^2\text{ gal}} \]

c) Mass of \(\text{CO}_2\) Released:

Reaction: \(\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)\)

Molar ratio is 1:1, so moles of \(\text{CO}_2 = 15,893.15\text{ mol}\).

\[ m = 15,893.15\text{ mol} \times 44.01\text{ g/mol} = 699,457\text{ g} \implies \mathbf{7.00 \times 10^2\text{ kg}} \]

d) Mass of \(\text{H}_2\text{O}\) Formed:

Molar ratio of \(\text{CH}_4\) to \(\text{H}_2\text{O}\) is 1:2.

\[ n = 2 \times 15,893.15 = 31,786.3\text{ mol} \]

\[ m = 31,786.3\text{ mol} \times 18.02\text{ g/mol} = 572,789\text{ g} \implies \mathbf{5.73 \times 10^2\text{ kg}} \]

e) Volume of Atmospheric Air Consumed:

Moles of \(\text{O}_2\) needed = \(2 \times 15,893.15 = 31,786.3\text{ mol}\).

Mass of \(\text{O}_2\) needed = \(31,786.3\text{ mol} \times 32.00\text{ g/mol} = 1,017,162\text{ g}\).

Total air mass required (air is 23% oxygen by mass):

\[ m_{\text{air}} = \frac{1,017,162\text{ g}}{0.23} = 4,422,443\text{ g} \]

Total air volume needed (density = \(1.22\text{ g/L}\)):

\[ V_{\text{air}} = \frac{4,422,443\text{ g}}{1.22\text{ g/L}} = 3,624,953\text{ L} \implies \mathbf{3.6 \times 10^6\text{ L}} \]

f) Electrical Equivalency Consumption:

\[ \text{Electrical work} = \mathbf{3500\text{ kWh}} \]

g) Raw Coal Mass Requisite (\(\text{kg}\)):

Step 1: Adjust electrical demand for power plant grid generation losses (40% efficiency):

\[ \text{Gross plant energy required} = \frac{3500\text{ kWh}}{0.40} = 8750\text{ kWh} \]

Step 2: Calculate mass of coal in pounds (yields \(2.26\text{ kWh/lb}\)):

\[ \text{Mass in lbs} = \frac{8750\text{ kWh}}{2.26\text{ kWh/lb}} = 3871.68\text{ lbs} \]

Step 3: Convert pounds to kilograms (\(1\text{ lb} = 0.453592\text{ kg}\)):

\[ m = 3871.68\text{ lbs} \times 0.453592\text{ kg/lb} = 1756.16\text{ kg} \implies \mathbf{1.76 \times 10^3\text{ kg}} \]