1. Chapter 6 Thermochemistry

2. Contents and Concepts Understanding Heats of Reaction
The first part of the chapter lays the groundwork
for understanding what we mean by heats of reaction.
Energy and Its Units
Heat of Reaction
Enthalpy and Enthalpy Changes
Thermochemical Equations
Applying Stoichiometry to Heats of Reaction
Measuring Heats of Reaction

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

4. Learning Objectives Understanding Heats of Reaction
Energy and Its Units
Define energy, kinetic energy, and internal energy.
Define the SI unit of energy (joule) as well as the common unit of energy (calorie).
Calculate the kinetic energy of a moving object.
State the law of conservation of energy.

5. Heat of Reaction
Define a thermodynamic system and its surroundings.
Define heat and heat of reaction.
Distinguish between an exothermic process and an endothermic process.
Enthalpy and Enthalpy Changes
Define enthalpy and enthalpy of reaction.
Explain how the terms enthalpy of reaction and heat of reaction are related.
Explain how enthalpy and internal energy are related.

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

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

8. Using Heats of Reaction
Hess’s Law
a. State Hess’s law of heat summation.
b. Apply Hess’s law to obtain the enthalpy change for one reaction from the enthalpy changes of a number of other reactions.

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

10. Fuels—Foods, Commercial Fuels, and Rocket Fuels
Define a fuel.
Describe the three needs of the body that are fulfilled by foods.
Give the approximate average values quoted (per gram) for the heat values (heats of combustion) for fats and for carbohydrates.
List the three major fossil fuels.
Describe the processes of coal gasification and coal liquefaction.
Describe some fuel-oxidizer systems used in rockets.

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

12. Energy
The potential or capacity to move matter.
One form of energy can be converted to another form of energy: electromagnetic, mechanical, electrical, or chemical.
Next, we’ll study kinetic energy, potential energy, and internal energy.

13. Kinetic Energy, EK
The energy associated with an object by virtue of its motion.
m = mass (kg)
v = velocity (m/s)

14. The SI unit of energy is the joule, J, pronounced “jewel.”
The calorie is a non-SI unit of energy commonly used by chemists. It was originally defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius. The exact definition is given by the equation:

15. A person weighing 75. 0 kg (165 lbs) runs a course at 1. 78 m/s (4
A person weighing 75.0 kg (165 lbs) runs a course at 1.78 m/s (4.00 mph). What is the person’s kinetic energy?
m = 75.0 kg
V = 1.78 m/s
EK = ½ mv2

16. g = gravitational constant (9.80 m/s2)
Potential Energy, EP
The energy an object has by virtue of its position in a field of force, such as gravitaitonal, electric or magnetic field.
Gravitational potential energy is given by the equation
m = mass (kg)
g = gravitational constant (9.80 m/s2)
h = height (m)

17. Internal Energy, U
The sum of the kinetic and potential energies of the particles making up a substance.
Total Energy
Etot = EK + EP + U

18. Law of Conservation of Energy
Energy may be converted from one form to another, but the total quantity of energy remains constant.

19. Thermodynamic System
The substance under study in which a change occurs is called the thermodynamic system (or just system).
Thermodynamic Surroundings
Everything else in the vicinity is called the thermodynamic surroundings (or just the surroundings).

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

21. Heat of Reaction
The value of q required to return a system to the given temperature at the completion of the reaction (at a given temperature)

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

23. In an endothermic reaction:
The reaction vessel cools.
Heat is absorbed.
Energy is added to the system.
q is positive.
In an exothermic reaction:
The reaction vessel warms.
Heat is evolved.
Energy is subtracted from the system.
q is negative.

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

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

26. The altitude of a campsite is a state function.
It is independent of the path taken to reach it.

27. DH = H(products) – H(reactants)
Enthalpy of Reaction
The change in enthalpy for a reaction at a given temperature and pressure:
DH = H(products) – H(reactants)
Note: D means “change in.”
Enthalpy change is equal to the heat of reaction at constant pressure:
DH = qP

28. 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
The diagram illustrates the enthalpy change for the reaction
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
The reactants are at the top. The products are at the bottom.
The products have less enthalpy than the reactants, so enthalpy is evolved as heat.
The signs of both q and DH are negative.

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

30. 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
For the reaction
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
–PDV
The H2 gas had to do work to raise the piston.
For the reaction as written at 1 atm, -PDV = -2.5 kJ.
In addition, kJ of heat are evolved.

31. Thermochemical Equation
The thermochemical equation is the chemical equation for a reaction (including phase labels) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation.
For the reaction of sodium metal with water, the thermochemical equation is:
2Na(s) + 2H2O(l) 
2NaOH(aq) + H2(g); DH = –368.6 kJ

32. Sulfur, S8, burns in air to produce sulfur dioxide
Sulfur, S8, burns in air to produce sulfur dioxide. The reaction evolves 9.31 kJ of heat per gram of sulfur at constant pressure. Write the thermochemical equation for this reaction.

33. S8(s) + 8O2(g)  8SO2(g); DH = –2.39 × 103 kJ
We first write the balanced chemical equation:
S8(s) + 8O2(g)  8SO2(g)
Next, we convert the heat per gram to heat per mole.
Note: The negative sign indicates that heat is evolved; the reaction is exothermic.
Now we can write the thermochemical equation:
S8(s) + 8O2(g)  8SO2(g); DH = –2.39 × 103 kJ

34. Manipulating a Thermochemical Equation
• When the equation is multiplied by a factor, the value of DH must be multiplied by the same factor.
• When a chemical equation is reversed, the sign of DH is reversed.

36. This reaction is identical to the given reaction.
a. CH4(g) + H2O(g)  CO(g) + 3H2(g)
This reaction is identical to the given reaction.
It is endothermic.
DH = 206 kJ
CH4(g) + H2O(g)  CO(g) + 3H2(g); DH = 206 kJ

37. This reaction is double the given reaction.
b. 2 CH4(g) + 2H2O(g)  2CO(g) + 6H2(g)
This reaction is double the given reaction.
It is endothermic.
DH = 412 kJ
CH4(g) + H2O(g)  CO(g) + 3H2(g); DH = 206 kJ

38. This reaction is the reverse of the given reaction.
c. CO(g) + 3H2(g)  CH4(g) + H2O(g)
This reaction is the reverse of the given reaction.
It is exothermic.
DH = -206 kJ
CH4(g) + H2O(g)  CO(g) + 3H2(g); DH = 206 kJ

39. d. 2CO(g) + 6H2(g)  2CH4(g) + 2H2O(g)
This reaction is reverse and double the given reaction.
It is exothermic.
DH = -412 kJ
Equations c and d are exothermic.
Equation d is the most exothermic reaction.
CH4(g) + H2O(g)  CO(g) + 3H2(g); DH = 206 kJ

40. When sulfur burns in air, the following reaction occurs:
S8(s) + 8O2(g)  8SO2(g);
DH = – 2.39 x 103 kJ
Write the thermochemical equation for the dissociation of one mole of sulfur dioxide into its elements.

41. S8(s) + 8O2(g)  8SO2(g); DH = –2.39 × 103 kJ
We want SO2 as a reactant, so we reverse the given reaction, changing the sign of DH:
8SO2(g)  S8(g) + 8O2(g) ; DH = × 103 kJ
We want only one mole SO2, so now we divide every coefficient and DH by 8:
SO2(g)  1/8S8(g) + O2(g) ; DH = +299 kJ

42. Applying Stoichiometry to Heats of Reaction

43. You burn 15.0 g sulfur in air. How much heat evolves from this amount of sulfur? The thermochemical equation is
S8(s) + 8O2(g)  8SO2(g); DH = x 103 kJ

44. S8(s) + 8O2(g)  8SO2(g); DH = -2.39 x 103 kJ
Molar mass of S8 = g
q = –1.40 × 102 kJ

45. The daily energy requirement for a 20-year-old man weighing 67 kg is 1
The daily energy requirement for a 20-year-old man weighing 67 kg is 1.3 x 104 kJ. For a 20-year-old woman weighing 58 kg, the daily requirement is
8.8 x 103 kJ. If all this energy were to be provided by the combustion of glucose, C6H12O6, how many grams of glucose would have to be consumed by the man and the woman per day?
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l);
DH = x 103 kJ

46. C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l); DH = -2.82 x 103 kJ
For a 20-year-old man weighing 67 kg:
= 830 g glucose required
(2 significant figures)
For a 20-year-old woman weighing 58 kg:
= 560 g glucose required
(2 significant figures)

47. Measuring Heats of Reaction
We will first look at the heat needed to raise the temperature of a substance because this is the basis of our measurements of heats of reaction.

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

49. The heat required can be found by using the following equations.
Using heat capacity:
q = CDt
Using specific heat capacity:
q = s x m x Dt

50. A calorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change. Two examples are shown below.

51. (3 significant figures)
A piece of zinc weighing 35.8 g was heated from 20.00°C to 28.00°C. How much heat was required? The specific heat of zinc is J/(g°C).
m = 35.8 g
s = J/(g°C)
Dt = 28.00°C – 20.00°C = 8.00°C
q = m  s  Dt
q = 111 J
(3 significant figures)

52. Nitromethane, CH3NO2, an organic solvent burns in oxygen according to the following reaction:
CH3NO2(g) + 3/4O2(g) 
CO2(g) + 3/2H2O(l) + 1/2N2(g)
You place g of nitromethane in a calorimeter with oxygen and ignite it. The temperature of the calorimeter increases from 22.23°C to 28.81°C. The heat capacity of the calorimeter was determined to be kJ/°C. Write the thermochemical equation for the reaction.

53. We first find the heat evolved for the 1.724 g of nitromethane, CH3NO2.
Now, covert that to the heat evolved per mole by using the molar mass of nitromethane, g.
DH = –709 kJ

54. CH3NO2(l) + ¾O2(g)  CO2(g) + 3/2H2O(l) + ½N2(g);
We can now write the thermochemical equation:
CH3NO2(l) + ¾O2(g)  CO2(g) + 3/2H2O(l) + ½N2(g);
DH = –709 kJ

55. Hess’s Law of Heat Summation
For a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps.

56. Suppose we want DH for the reaction 2C(graphite) + O2(g)  2CO(g)
It is difficult to measure directly.
However, two other reactions are known:
2CO2(g)  2CO(g) + O2(g); DH = – kJ
C(graphite) + O2(g)  CO2(g); DH = kJ
In order for these to add to give the reaction we want, we must multiply the first reaction by 2. Note that we also multiply DH by 2.

57. 2C(graphite) + O2(g)  2CO(g)
2C(graphite) + 2O2(g)  2CO2(g); DH = kJ
2CO2(g)  2CO(g) + O2(g); DH = – kJ
2 C(graphite) + O2(g)  2 CO(g); DH = – kJ
Now we can add the reactions and the DH values.

58. DHsub = DHfus + DHvap

59. What is the enthalpy of reaction, DH, for the reaction of calcium metal with water?
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
This reaction occurs very slowly, so it is impractical to measure DH directly. However, the following facts are known:
H+(aq) + OH-(aq)  H2O(l); DH = –55.9 kJ
Ca(s) + 2H+(aq)
 Ca2+(aq) + H2(g); DH = –543.0 kJ

60. Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
First, identify each reactant and product:
H+(aq) + OH-(aq)  H2O(l); DH = –55.9 kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
Each substance must be on the proper side.
Ca(s), Ca2+(aq), and H2(g) are fine.
H2O(l) should be a reactant.
OH-(aq) should be a product.
Reversing the first reaction and changing the sign of its DH accomplishes this.

61. Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
H2O(l)  H+(aq) + OH-(aq); DH = kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
The coefficients must match those in the reaction we want.
The coefficient on H2O and OH- should be 2. We multiply the first reaction and its DH by 2 to accomplish this.

62. Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
2H2O(l)  2H+(aq) + 2OH-(aq); DH = kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
We can now add the equations and their DH’s.
Note that 2H+(aq) appears as both a reactant and a product.

63. Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
2H2O(l)  2H+(aq) + 2OH-(aq); DH = kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g);
DH = –431.2 kJ

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

65. Elements can exist in more than one physical state, and some elements exist in more than one distinct form in the same physical state. For example, carbon can exist as graphite or as diamond; oxygen can exist as O2 or as O3 (ozone).
These different forms of an element in the same physical state are called allotropes.
The reference form is the most stable form of the element (both physical state and allotrope).

66. The standard enthalpy of formation, DHf°, is the enthalpy change for the formation of one mole of the substance from its elements in their reference forms and in their standard states. DHf° for an element in its reference and standard state is zero.
For example, the standard enthalpy of formation for liquid water is the enthalpy change for the reaction
H2(g) + 1/2O2(g)  H2O(l)
DHf° = –285.8 kJ
Other DHf° values are given in Table 6.2 and Appendix C.

67. CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
1/2 H2(g) + 1/2 Cl2(g)  HCl(g); DHf° = kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = kJ
We first identify each reactant and product from the reaction we want.

68. CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
1/2 H2(g) + 1/2 Cl2(g)  HCl(g); DHf° = kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = kJ
Each needs to be on the correct side of the arrow and is. Next, we’ll check coefficients.

69. CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
1/2 H2(g) + 1/2 Cl2(g)  HCl(g); DHf° = kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = kJ
Cl2 and HCl need a coefficient of 4. Multiplying the second equation and its DH by 4 does this.

70. CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
2H2(g) + 2Cl2(g)  4HCl(g); DHf° = kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = kJ
Now, we can add the equations.

71. CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
2H2(g) + 2Cl2(g)  4HCl(g); DHf° = kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = kJ
CH4(g) + 4Cl2(g)  CCl4(g) + 4HCl(g); DH° = –429.7 kJ

72. What is the heat of vaporization of methanol, CH3OH, at 25°C and 1 atm?
Use standard enthalpies of formation (Appendix C).

73. We want DH° for the reaction:
CH3OH(l)  CH3OH(g)
DHvap= kJ

74. Methyl alcohol, CH3OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO, which can coagulate protein. Calculate DHo for the following reaction:
2CH3OH(aq) + O2(g)  2HCHO(aq) + 2H2O(l)
Standard enthalpies of formation, :
CH3OH(aq): kJ/mol
HCHO(aq): kJ/mol
H2O(l): kJ/mol

75. 2CH3OH(aq) + O2(aq)  2HCHO(aq) + 2H2O(l)
We want DH° for the reaction:
2CH3OH(aq) + O2(aq)  2HCHO(aq) + 2H2O(l)

76. Foods fuels three needs of the body:
• They supply substances for the growth and repair of tissue.
• They supply substances for the synthesis of compounds used in the regulation processes.
• They supply energy.
Foods do this by a combustion process.

77. For glucose, a carbohydrate:
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l); DHf° = –2803 kJ
For glycerol trimyristate, a fat:
C45H86O6(s) + 127/2O2(g)  45CO2(g) + 43H2O(l); DHf°= –27,820 kJ
The average value for carbohydrates is 4.0 kcal/g and for fats is 9.0 kcal/g.

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

79. Coal, which accounts for 22. 9% of total U. S
Coal, which accounts for 22.9% of total U.S. energy consumption, varies in terms of the amount of carbon it contains and so varies in terms of the amount of energy it produces in combustion.
Anthracite (hard coal) was laid down as long as 250 million years ago and can contain as much as 80% carbon. Bituminous coal, a younger variety, contains between 45% and 65% carbon.

80. Natural gas, which accounts for 22. 7% of total U. S
Natural gas, which accounts for 22.7% of total U.S. energy consumption, is convenient because it is fluid and can be easily transported. Purified natural gas is primarily methane, CH4, plus small amounts of ethane, C2H6; propane, C3H8; and butane, C4H10.
Petroleum is a mixture of compounds. Gasoline, which is obtained from petroleum, is a mixture of hydrocarbons (compounds of carbon and hydrogen).

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

82. Coal gasification is one way. Steam is passed over hot coal:
C(s) + H2O(g)  CO(g) + H2(g)
The mixture containing carbon monoxide can be converted by various methods into useful products.
The mixture of carbon monoxide and hydrogen can be converted by various methods into useful products such as methane, CH4.
CO(g) + 3H2(g)  CH4(g) + H2O(g)

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