ANSWERS TO
END-OF-CHAPTER QUESTIONS
CHAPTER 4: ENERGY, CHEMISTRY, AND SOCIETY
Emphasizing Essentials
1. a. List three fossil fuels.
b. What is the origin of fossil fuels?
c. Are fossil fuels a renewable resource?
Answer:
a. Coal, oil, and natural gas.
b. Fossil fuels originated hundreds of millions of years ago from ancient plant and animal
matter. Over hundreds of millions of years, these materials were transformed into coal,
petroleum, and natural gas by the action of heat and pressure.
c. No. The length of time required to form these fuels means they are not renewable.
2. Explain each energy transformation step that takes place when coal is burned in a power plant.
Answer:
Coal is a source of potential energy. When burned, some of its potential energy is converted to
heat through combustion. The heat is converted into kinetic energy of the vaporized water
molecules (steam). The kinetic energy is converted to mechanical energy by spinning a
turbine. The mechanical energy generated from the spinning turbine is converted to electrical
energy by rotating a wire in a magnetic field.
3. Compare the processes of combustion and photosynthesis.
Answer:
The processes of combustion (for carbon-based fuels) and photosynthesis (for green plants)
are both involve CO2 and H2O, but for the former these compounds are products and for the
latter they are reactants. Combustion takes fuels with high potential energy and converts them
to products with low potential energy (CO2 and H2O). Photosynthesis is the chemical reaction
of CO2 and H2O to produce O2 and glucose, substances of high potential energy. The
processes also differ in that photosynthesis requires the input of energy (in the form of
sunlight) and combustion releases energy (in several forms, including heat and light).
4. Describe how grades of coal differ and the significance of these differences.
Answer:
Coal is typically rated on the percentages of carbon and oxygen it contains. High grade coal
(e.g., anthracite, bituminous) has a relatively high percentage of carbon and is also low in
sulfur. Low grade coal (e.g., lignite) has a higher oxygen and moisture content and has an
energy content only slightly higher than wood. Higher grade coals release more energy per
gram and produce less sulfur dioxide.
PAGE 4-1
5. A coal-burning power plant generates electrical power at a rate of 500 megawatts (MW)
(5.00 × 108 J/s). The plant has an overall efficiency of 37.5% (0.375) for the conversion of heat
to electricity.
a. Calculate the electrical energy (in joules) generated in 1 year of operation and the heat
energy used for that purpose.
b. Assuming the power plant burns coal that releases 30 kJ/g, calculate the mass of coal (in
grams and metric tons) that is burned in 1 year of operation. Hint: 1 metric ton = 1 × 103 kg
= 1 × 106 g.
Answer:
5.00 x 10 8 J 3600 s 24 hr 365 days
a.
= 1.58  1016 J of electricity generated per year



s
hr
day
year
1.58 ×1016 J = 4.2  1016 J of heat for electricity generation
0.375
b. 4.2 × 1016 J ×
1 kJ
1g
= 1.4  1012 g

1000J 30 kJ
1.4 × 1012 g ×
1000 kg
10 6 g
= 1.4  109 metric tons
6. The complete combustion of methane is given in equation 4.3.
a. By analogy, write a similar chemical equation using ethane, C2H6.
b. Rewrite this equation with Lewis structures.
c. The heat of combustion for ethane, C2H6, is 52.0 kJ/g. How much heat is if 1.0 mol of
ethane undergoes complete combustion?
Answer:
a.
b.
c. 1.0 mol C2H6 
30 g C 2 H 6
52.0 kJ

= 1560 kJ/mol C2H6
1 mol C 2 H 6 mol C 2 H 6
7. List three major drawbacks of using coal as a fuel.
Answer:
Coal is difficult to transport and dangerous to mine. Coal is also a dirty fuel that releases
PAGE 4-2
mercury in its fly ash, and contributes to acid rain by releasing NOx and SOx to the
atmosphere when it is burned. In addition, coal combustion contributes to global warming
by releasing CO2 into the atmosphere.
8. Mercury (Hg) is a contaminant of coal, ranging from 50–200 ppb. Consider the amount of coal
burned by the power plant in Your Turn 4.8. Calculate tons of mercury in the coal based on
the lower (50 ppb) and higher (200 ppb) concentrations.
Answer:
A typical power plant burns 1.5 million tons of coal each year. The first calculation is for
coal with 50 ppb mercury; the second is for 200 ppb.
x ton Hg
50 ton Hg

6
1.5  10 ton coal 1  109 toncoal
x ton Hg
200 ton Hg

6
1.5  10 ton coal
1  109 ton coal

x = 0.075 ton Hg
x = 0.30 ton Hg
Assuming mercury concentrations in the range of 50-200 ppb, the plant releases between
0.075 and 0.30 tons of Hg per year.
9. An energy consumption of 650,000 kcal per person per day is equivalent to an annual
consumption of 65 barrels of oil or 16 tons of coal. Calculate the amount of energy available
in kilocalories for each of these.
a. one barrel of oil
b. 1 gallon of oil (42 gallons per barrel)
c. 1 ton of coal
d. 1 pound of coal (2000 pounds per ton)
Answer:
650,000 kcal 365 day 2.4  108 kcal


1 day
1 yr
1 yr
This value can be related to each of the energy sources.
a.
1 yr
3.7  10 6 kcal
2.4  10 8 kcal


1 yr
65 barrel
barrel
b.
2.4  108 kcal
1 yr
1 barrel oil 8.8  10 4 kcal


=
1 yr
65 barrel oil
42 gal
1 gal
2.4  108 kcal
1 yr
1.5  10 7 kcal

=
1 yr
16 ton coal
1 ton coal
7
1.5  10 kcal
1 ton
7.5  10 3 kcal
d.


1 ton
2000 pound
pound
c.
10. Use the information in the previous question to find the ratio of the quantity of energy
PAGE 4-3
available in 1 pound of coal to that in 1 pound of oil. Hint: One pound of oil has a volume of
0.56 quart.
Answer:
8.8  10 4 kcal 1 gallon 0.56 quarts 1.2  10 4 kcal



1 gallon oil 4 quarts 1 pound oil
1 pound oil
1.2 10 4 kcal 1 pound coal

 1.6
1 pound oil 7.5 10 3 kcal
A pound of oil has 1.6 times as much energy available as a pound of coal.
11. Consider the data for these three hydrocarbons.
Compound, Formula Melting Point
Boiling
Point
(°C)
(°C)
pentane, C5H12
-130
36
triacontane, C30H62
66
450
octane, C8H18
-57
125
Predict the physical state (solid, liquid, or gas) of each at room temperature.
Answer:
Assuming that room temperature is 20-25 °C, pentane should be a liquid because room
temperature is below its boiling point (36 °C) but above its melting point (-130 °C).
Triacontane should be solid at room temperature because room temperature is below its
melting point (66 °C). Octane should be a liquid at room temperature for the same reason as
pentane.
12. a. Write the chemical equation for the complete combustion of heptane, C7H16.
b. The heat of combustion for heptane is 4817 kJ/mol. How much heat is released if 250 kg
of n-heptane burns completely?
Answer:
a. C7H16 + 11 O2 → 7 CO2 + 8 H2O
10 3 g C7 H16 1 mol C 7 H16
4817 kJ


 1.2  10 7 kJ
b. 250 kg C7 H16 
kg C 7 H16
100.2 g C 7 H16 1 mol C 7 H16
13. Figure 4.17 shows energy differences for the combustion of hydrogen, an exothermic
chemical reaction. The combination of nitrogen gas and oxygen gas to form nitrogen

monoxide is an example of an endothermic reaction:
The bond energy in NO is 630 kJ/mol. Sketch an energy diagram for this reaction, and
calculate the overall energy change.
Answer:
Note that in an endothermic reaction, the potential energy of the products is greater than the
potential energy of the reactants. The opposite was true for an exothermic reaction.
PAGE 4-4
Breaking 1 mole
O=O double bonds
= +498 kJ
Forming 2 moles
N=O double bonds
= 2 x (–630 kJ)
= –1260 kJ
Breaking 1 mole
N≡N triple bonds
= +946 kJ
2NO (products)
N2 + O2
Net energy change
= 946 + 498 – 1260 = +184 kJ
(reactants)
14. A single serving bag of Granny Goose Hawaiian Style Potato Chips has 70 Cal. Assuming
that all of the energy from eating these chips goes toward keeping your heart beating, how
long can these chips sustain a heartbeat of 80 beats per minute? Note: 1 kcal 5 4.184 kJ, and
each human heart beat requires approximately 1 J of energy.
Answer:
70 Cal 
4.184 kJ 1000 J 1 beat
1 min
= 3700 min



1 Cal
1 kJ
1J
80 beats
15. A 12-oz serving of a soft drink has an energy equivalent of 92 kcal.
a. In kilojoules, what is the energy released when metabolizing this beverage?
b. Assume that you use this energy to lift concrete blocks that weigh 22 lb (10 kg) each. How
many blocks could you lift to a height of 4 ft with the energy calculated in part a.
PAGE 4-5
Answer:
4.184 kJ
= 380 kJ
1 kcal
100 cm
22 lb
1 kJ
1J
b. 3 m 



= 0.66 kJ/block
1m
10 cm 1000 J 1 block
a. 92 kcal 
380 kJ 
0.66 kJ
= 250 blocks
1 block
16. One way to produce ethanol for use as a gasoline additive is the reaction of water vapor with
ethylene:
a. Rewrite this equation using Lewis structures.
b. Is this reaction endothermic or exothermic?
c. In your calculation, was it necessary to break all the chemical bonds in the reactants to
form the product ethanol? Explain your answer.
Answer:
a.
b. Bonds broken in the reactants:
1 mol C–C double bond = 1(598 kJ)
= 598 kJ
1 mol O–H single bond = 1(467 kJ)
= 467 kJ
Total energy absorbed in breaking bonds = 1065 kJ
Bonds formed in the products:
1 mol C–C single bond = 1(356 kJ)
= 356 kJ
1 mol C–H single bond = 1(416 kJ)
= 416 kJ
1 mol C–O single bond = 1(336 kJ)
= 336 kJ
Total energy released in forming bonds = 1108 kJ
Net energy change: (1065 kJ) + (1108 kJ) = – 43 kJ
Because the energy released in forming bonds is greater than the energy absorbed in breaking
bonds, the net energy change is negative and the overall reaction is exothermic.
c. No, it is not necessary to break all of the bonds. There are four carbon-to-hydrogen single
bonds on both the reactant side and the product side. Similarly, there is one O–H bond on
both sides.
17. From personal experience, state whether these processes are endothermic or exothermic.
Give a reason for each.
a. A charcoal briquette burns.
b. Water evaporates from your skin.
PAGE 4-6
c. Ice melts.
Answer:
a. Exothermic. A charcoal briquette releases heat as it burns.
b. Endothermic. Water absorbs the heat necessary for evaporation from your skin, and your
skin feels cooler.
c. Endothermic. Ice absorbs the necessary heat to melt from the environment.
18. Use the bond energies in Table 4.4 to explain why:
a. chlorofluorocarbons, CFCs, are so stable.
b. it takes less energy to release Cl atoms than F atoms from CFCs.
Answer:
a. CFCs are stable because the bond energies for C–Cl and C–F are high compared to other
bond energies.
b. The C–Cl bond energy (327 kJ/mol) is lower than the C–F bond energy (485 kJ/mol).
19. Use the bond energies in Table 4.4 to calculate the energy changes associated with each of
these reactions. Label each reaction as endothermic or exothermic. Hint: Draw Lewis
structures of the reactants and products to determine the number and kinds of bonds.
a.
b.
c.
Answer:
a. Bonds broken in the reactants
1 mol N≡N triple bonds
= 1(946 kJ)
3 mol H–H single bonds
= 3(436 kJ)
Total energy absorbed in breaking bonds
= 946 kJ
= 1308 kJ
= 2254 kJ
Bonds formed in the products
6 mol N–H single bonds
= 6(391 kJ)
Total energy released in forming bonds
= 2346 kJ
= 2346 kJ
Net energy change is (+2254 kJ) + (2346 kJ)
= 92 kJ
The overall energy change is negative, characteristic of an exothermic reaction.
b. Bonds broken in the reactants
12 mol C–H single bonds
= 12(416 kJ)
4 mol C–C single bonds
= 4(356 kJ)
11 mol O=O double bonds
= 11(498 kJ)
Total energy absorbed in breaking bonds
= 4,992 kJ
= 1424 kJ
= 5478 kJ
= 11,894 kJ
Bonds formed in the products
10 mol C≡O triple bonds
= 10,730 kJ
= 10(1073 kJ)
PAGE 4-7
24 mol O–H single bonds
= 24(467 kJ)
Total energy released in forming bonds
= 11,208 kJ
= 21,938 kJ
Net energy change is (+11,894 kJ) + (21,938 kJ)
= 10,044 kJ
The overall energy change is negative, characteristic of an exothermic reaction.
c. Bonds broken in the reactants
1 mol H–H single bonds
= 1(436 kJ)
1 mol Cl–Cl single bonds
= 1(242 kJ)
Total energy absorbed in breaking bonds
= 436 kJ
= 242 kJ
= 678 kJ
Bonds formed in the products
2 mol H–Cl single bonds
= 2(431 kJ)
Total energy released in forming bonds
= 862 kJ
= 862 kJ
Net energy change is (+678 kJ) + (862 kJ)
= 184 kJ
The overall energy change is negative, characteristic of an exothermic reaction.
20. Use the bond energies in Table 4.4 to calculate the energy changes associated with each of
these reactions. Label each reaction as endothermic or exothermic.
a.
b.
c.
Answer:
a. Bonds broken in the reactants
2 mol H–H single bonds
= 2(436 kJ) = 872 kJ
1 mol C≡O triple bonds
=1(1073 kJ) =1073 kJ
Total energy absorbed in breaking bonds =1945 kJ
Bonds formed in the products
1 mol C–O single bonds
=1(336 kJ)
1 mol O–H single bonds
=1(467 kJ)
3 mol C–H single bonds
=3(416 kJ)
Total energy released in making bonds
= 336 kJ
= 467 kJ
=1248 kJ
=2051 kJ
Net energy change is (+1945 kJ) + (2051 kJ) = 56 kJ
The overall energy change is negative, characteristic of an exothermic reaction.
b. Bonds broken in the reactants
1 mol H–H single bonds
= 1(436 kJ)
1 mol O=O double bonds
= 1(498 kJ)
Total energy absorbed in breaking bonds
= 436 kJ
= 498 kJ
= 934 kJ
Bonds formed in the products
PAGE 4-8
2 mol O–H single bonds
= 2(467 kJ)
1 mol O–O single bonds
1(146 kJ)
Total energy released in making bonds
= 934 kJ
= 146 kJ
= 1080 kJ
Net energy change is (+934 kJ) + (1080 kJ)
= 146 kJ
The overall energy change is zero, indicating the reaction is neither endothermic nor
exothermic.
c. Bonds broken in the reactants
2 mol Br–Cl single bonds
= 2(217 kJ)
Total energy absorbed in breaking bonds
= 434 kJ
= 434 kJ
Bonds formed in the products
1 mol Br–Br single bonds
= 1(193 kJ)
= 193 kJ
1 mol Cl–Cl single bonds
= 1(242 kJ)
= 242 kJ
Total energy released in making bonds
= 435 kJ
Net energy change is (+434 kJ) + (435 kJ)
= 1 kJ
The overall energy change is negative, characteristic of an exothermic reaction.
21. Use Figure 4.6 to compare the sources of U.S. energy consumption. Arrange the sources in
order of decreasing percentage and comment on the relative rankings.
Answer:
Petroleum > Natural Gas ~ Coal > Nuclear > Hydro ~ Wood. The three largest energy
sources are non-renewable fossil fuels. The renewable sources, hydro and wood, are the
smallest contributors, and solar and wind power are not yet visible on the graph.
22. The structural formulas of straight-chain (normal) alkanes containing 1 to 8 carbon atoms are
given in Table 4.2.
a. Draw the structural formula for n-decane, C10H22.
b. Predict the chemical formula for n-nonane (9 carbon atoms) and for n-dodecane (12
carbon atoms).
c. The structural formulas shown are two-dimensional. Use the bond angle information in
Chapter 3 to predict the C–C–C and H–C–H bond angles in n-decane.
Answer:
a. Decane:
H
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
PAGE 4-9
H
b. nonane:
dodecane:
c. In these molecules, each carbon atom forms four single bonds, either to other carbon
atoms or to hydrogen atoms. Thus the geometry around the each carbon atom is tetrahedral
and the bond angles are approximately 109.5o.
23. Consider this equation representing the process of cracking.
a. Which bonds are broken and which bonds are formed in this reaction? Use Lewis
structures to help answer this question.
b. Use the information from part a and Table 4.4 to calculate the energy change during this
cracking reaction.
Answer:
a.
b. Bonds broken in the reactants
1 mol C–C single bonds
= 1(356 kJ)
1 mol C–H single bonds
= 1(416 kJ)
Total energy absorbed in breaking bonds
= 356 kJ
= 416 kJ
= 772 kJ
Bonds formed in the products
1 mol C–H single bonds
1 mol C=C double bonds
Total energy released in forming bonds
= 416 kJ
= 598 kJ
=1014 kJ
= 1(416 kJ)
= 1(598 kJ)
Net energy change is (+772 kJ) + (–1014kJ) = –242 kJ
The overall energy change has a negative sign, characteristic of an exothermic reaction.
24. Here is a ball-and-stick representation for one isomer of butane (C4H10).
PAGE 4-10
a. Draw the Lewis structure for this isomer.
b. Draw Lewis structures for all other isomers. Hint: Watch for duplications!
Answer:
a.
b.
25. A premium gasoline available at most stations has an octane rating of 92. What does that tell
you about:
a. the knocking characteristics of this gasoline?
b. whether the fuel contains oxygenates?
Answer:
a. Gasoline with an octane rating of 92 has the same knocking characteristics as a mixture
composed of 92% isooctane and 8% n-heptane. As a “premium gasoline,” it has a higher
octane rating than other blends sold at gasoline stations and hence is more resistant to
knocking than these blends. Note: check the owner’s manual to determine if this blend is
appropriate.
b. The octane rating provides no information about whether or not the fuel contains
oxygenates. Other labels around the pump should reveal this information.
26.
Figure 4.16 gives the energy content of several fuels in kilojoules per gram (kJ/g).
Calculate the energy content in kilojoules per mole (kJ/mol) for each. How does the
chemical composition of a fuel relate to its energy content? Visit Figures Alive! at the
textbook’s website for related activities.
Answer:
Hydrogen:
140 kJ 2.02 g

= 280 kJ/mol
1g
1 mol
PAGE 4-11
Methane:
56 kJ 16.05 g

= 900 kJ/mol
1 mol
1g
Propane:
51 kJ 44.11 g

= 2.2  103 kJ/mol
1 mol
1g
Gasoline (octane):
48 kJ 114 g

= 5.5  103 kJ/mol
1 mol
1g
Coal ( C135H96O9NS):
Ethanol (C2H6O):
Wood: C6H12O6:
31 kJ 1908 g

= 5.9  104 kJ/mol
1 mol
1g
30 kJ 46 g

= 1.4  103 kJ/mol
1 mol
1g
14.2 kJ 180 g

= 2.6  103 kJ/mol
1 mol
1g
As the oxygen content of the fuels increases, the energy released per gram decreases. For
hydrocarbons, as the H/C ratio increases, the energy released per gram increases.
Concentrating on Concepts
27. How might you explain the difference between temperature and heat to a friend? Use some
practical, everyday examples.
Answer:
Wouldn’t you rather spill a drop of hot coffee on you than the whole cupful at the same
temperature? Although the drop and the cup full of coffee may initially have the same
temperature, you will receive a bigger burn from the bigger volume of coffee because it has
the higher heat content. Heat is a form of energy. In contrast, temperature is a measurement
that indicates the direction heat will flow. Heat always flows from an object at high
temperature to an object at lower temperature. This means that if hot coffee is added to cold
coffee, heat will flow from the hot liquid to the cold liquid, and the final temperature of the
mixture will be between the original temperatures of the two individual solutions. Heat
depends on the temperature and on how much material is present.
28. Write a response to this statement: “Because of the first law of thermodynamics, there can
never be an energy crisis.”
Answer:
The first law of thermodynamics states that energy is neither created nor destroyed; it only
changes form. Energy can be transformed, but the total energy in the world is constant.
PAGE 4-12
However, this statement does not take into effect our ability to capture and use energy in all
of its forms. The energy of fossil fuels is stored in the form of chemical bonds. When we
burn fossil fuels, we release some of the energy stored Wind and solar power derive their
energy from the Sun. An energy crisis arises when demand exceeds supply. Unless we are
better able to capture energy from sources other than fossil fuels, we will indeed have an
energy crisis – not due to a shortage of absolute energy but a shortage in our ability to use the
energy available.
29. A friend tells you that hydrocarbon fuels containing larger molecules liberate more heat than
those with smaller ones.
a. Use these data, together with appropriate calculations, to discuss the merits of this
statement.
Hydrocarbon
Heat of Combustion
octane, C8H18
5450 kJ/mol
butane, C4H10
2859 kJ/mol
b. Based on your answer to part a, do you expect the heat of combustion per gram of candle
wax, C25H52, to be more or less than that of octane? Do you expect the molar heat of
combustion of candle wax to be more or less than that of octane? Justify your predictions.
Answer:
a. Considering only the molar heats of combustion, octane, with more atoms and more
chemical bonds, has a greater heat of combustion than butane. However, comparisons should
be based on the same amount of each substance, such as the heat released per gram of each
fuel.
5450 kJ
1 mol C 8 H18
47.7 kJ


, the heat released per gram octane burned.
1 mol C 8 H18 114.2 g C 8 H18 g C8 H18


2859 kJ
1 mol C 4 H10 49.2 kJ
, the heat released per gram butane burned.


1 mol C 4 H10 58.1 g C 4 H10 g C 4 H10
Here the values are much closer and with just two data points, it is not possible to establish a
trend. Notice, however, that the smaller hydrocarbon releases slightly more heat per gram
than the larger one. Another piece of supporting evidence is the heat of combustion of
methane, the smallest hydrocarbon, is even larger, 54 kJ/g. Because heat comparisons should
be made based on the same mass of fuel, you will have to educate your friend on this point.
b. Candle wax is composed of high molar mass hydrocarbons. Looking at the values from
part a, the heat of combustion per gram is expected to be slightly smaller and the heat of
combustion per mole is expected to be larger.
30. Halons are synthetic chemicals similar to CFCs but include bromine. Although halons are
excellent materials for fire fighting, they more effectively deplete ozone than CFCs. Here is
the Lewis structure for halon-1211.
PAGE 4-13
a. Which bond in this compound is broken most easily? How is that related to the ability of
this compound to deplete ozone?
b. The compound C2HClF4 is being considered as a replacement for halons in fire
extinguishers. Draw its Lewis structure and identify the bond broken most easily.
Answer:
a. The C–F single bond requires 485 kJ/mol, the C–Cl single bond requires 327 kJ/mol, and
the C–Br single bond requires 285 kJ/mol to break the bond. The C–Br bond is the weakest.
Thus, when halon-1211 absorbs UV radiation, bromine atoms are likely to form and react
with ozone.
b.
In this molecule, the C–Cl bond has the lowest bond energy and thus is broken most easily.
31. The Fischer–Tropsch conversion of hydrogen and carbon monoxide into hydrocarbons and
water was given in equation 4.11:
a. Determine the heat evolved by this reaction when n = 1.
b. Without doing a calculation, do you think that more or less energy is given off per mole in
the formation of larger hydrocarbons (n > 1)? Explain your reasoning.
Answer:
a. When n = 1, the balanced equation is
To calculate the heat evolved we use the same method as in Problem 4.16.
Bonds broken in the reactants
1 mol C≡O triple bonds
= 1(1073 kJ)
3 mol H–H single bonds
= 3(436 kJ)
Total energy absorbed in breaking bonds
= 1073 kJ
= 1308 kJ
= 2381 kJ
Bonds formed in the products
4 mol C–H single bonds
2 mol O–H single bonds
= 1664 kJ
= 934 kJ
= 4(416 kJ)
= 2(467 kJ)
PAGE 4-14
Total energy released in forming bonds
= 2598 kJ
Net energy change is (+2381 kJ) + (2598 kJ)
= 217 kJ
b. Reactions with n greater than 1 will release more energy as n becomes larger, assuming
that we are viewing the energy per mole of the hydrocarbon formed (not per gram). There
will always be n C≡O triple bonds to break and (2n + 1) H–H single bonds to break. The
number of C–H bonds forming will be (2n + 2), and the number of O–H bonds forming is
2n. As n becomes larger, more and more energy will be released.
32. During the distillation of petroleum, kerosene and hydrocarbons with 12–18 carbons used for
diesel fuel will condense at position C marked on this diagram.
a. Separating hydrocarbons by distillation depends on differences in a specific physical
property. Which one?
b. How does the number of carbon atoms in the hydrocarbon molecules separated at A, B,
and D compare with those separated at position C? Explain your prediction.
c. How do the uses of the hydrocarbons separated at A, B, and D differ from those separated
at position C? Explain your reasoning.
Answer:
a. Hydrocarbons separate due to differences in their boiling points.
b. Hydrocarbons separated at positions A and B have lower boiling points and are more
volatile than the hydrocarbons separated at position C. The hydrocarbons separated at
positions A and B have fewer carbons in their structures than those separated at position C.
The hydrocarbons at position D will be less volatile or not volatile at all, compared to those
at position C. The hydrocarbons separated at position D have more carbon atoms than the
hydrocarbons separated at C.
c. The hydrocarbons separated at A will be gases, and can be used as fuels and starting
materials for manufacturing. Those separated at B will be liquids, and can be used as motor
fuels and as industrial solvents. Position D contains residue material that is rich in many
complex compounds as well as many hydrocarbons. In addition to waxes and asphalt, these
tars can be further separated into other useful compounds. The hydrocarbons separated at C
are used as kerosene or diesel fuel or may be cracked.
33. Explain why cracking is necessary in the refinement of crude oil.
Answer:
Cracking is necessary because the demand for the mid-range hydrocarbons (C7-C12) found
PAGE 4-15
in gasoline exceeds the amount produced by the distillation of crude oil.
34. Consider Equation 4.15. Are glycerol and propylene glycol isomers? Explain.
Answer:
Glycerol (C3H8O3) and propylene glycol (C3H8O2) are not isomers because they have
different chemical formulas.
35. Catalysts speed up cracking reactions in oil refining and allow them to be carried out at lower
temperatures. What other examples of catalysts were given in the first three chapters of this
text?
Answer:
Section 1.11 described the catalytic converters in automobiles. Section 2.9 described the
catalytic destruction of ozone by chlorine free radicals.
36. Octane ratings of several substances are listed in Table 4.5.
a. What evidence can you give that the octane rating is or is not a measure of the energy
content of a gasoline?
b. Octane ratings are measures of a fuel’s ability to minimize or prevent engine knocking.
Why is the prevention of knocking important?
c. Why are higher octane gasolines more expensive than lower octane gasoline?
Answer:
a. Table 4.5 shows that both octane and isooctane have nearly identical heats of combustion.
These two compounds have the same chemical formula and release the same energy when
burned, so the octane rating cannot be a measure of the energy content of a gasoline.
b. Knocking produces an objectionable pinging sound, reduced engine power, overheating,
and possible engine damage.
c. The higher octane gasolines are more expensive to produce because they require more
“processing”, including energy-intensive cracking reactions that convert lower octane
molecules into higher octane molecules.
37. Section 4.8 states that both n-octane and isooctane have essentially the same heat of
combustion. How is that possible if they have different structures?
Answer:
Isooctane and n-octane are isomers, that is, they have the same chemical formula. When
these compounds are burned, the same number and type of bonds will be broken and formed.
Therefore, the heats of combustion are essentially the same.
38. One consequence of the heavy dependence of the United States on foreign oil is possible
periodic gasoline shortages due to international events. Gasoline shortages affect more than
individual motorists. List some ways in which a gasoline shortage could affect your life.
Answer:
PAGE 4-16
A gasoline shortage would affect far more than just individual motorists. For example,
gasoline is needed to fuel many commercial vehicles such as ambulances, police cars, and
some trucks. Gasoline also fuels the engines in lawn mowers and tractors. Higher fuel costs
are likely to be passed along to consumers.
39. It was stated in the text that emissions of some pollutants are lower using biodiesel than
using petroleum diesel. Based on the methods of production for each fuel, explain the lower
amounts of
a. sulfur dioxide emissions.
b. CO emissions.
Answer:
a. Biodiesel is produced from vegetable oil and fat so it does not contain sulfur.
b. In contrast to petroleum diesel, vegetable oil and fats are compounds that contain oxygen.
At least in theory, these molecules should burn more completely than non-oxygenated
compounds, thus producing less CO. In fact, this may not always be the case, as the
conditions under which a fuel is burned also determine how much CO is produced.
40. These three structures have the chemical formula C8H18. The hydrogen atoms and C–H bonds
have been omitted for simplicity.
a. Redraw the structures to show the missing hydrogen atoms. Hint: Check that all structures
have 18 H atoms.
b. Which (if any) of these structures are identical?
c. Obtain a model kit and build one of these molecules. What are the C–C–C bond angles?
d. Draw the structural formulas of two additional isomers of C8H18.
e. If you were to build models of these two isomers, would the C–C–C bond angles be the
same as that in part c? Explain.
Answer:
a. Each structure has the formula C8H18.
b. Yes; structures 2 and 3 have exactly the same order of linkage.
c. Each C–C–C bond angle is 109.5° because the geometry of the bonds around each carbon
atom is tetrahedral.
d. Several other isomers are possible. Be sure the linkage is different from the given isomers.
e. The bond angles are the same as the structures in part c. In every isomer of C8H18, each
carbon atom has four bonds and therefore the geometry is tetrahedral.
41. Here is a ball-and-stick model of ethanol, C2H6O. Another compound, dimethyl
PAGE 4-17
ether, has this same chemical formula. Draw the Lewis structure of diethyl ether.
Hint: Remember to follow the octet rule.
Answer:
The Lewis structure for dimethyl ether is:
42. Describe how the growth in oxygenated gasolines relates to each of these.
a. restrictions on the use of lead in gasoline.
b. federal and state air quality regulations.
Answer:
a. Eliminating the use of tetraethyl lead as an octane booster in part led to the rise of
oxygenates as octane boosters.
b. Cities in the U.S. that do not meet federal air quality standards are required by the Clean
Air Act of 1990 to use oxygenated fuels. The Environmental Protection Agency (EPA) web
site has more information on the Clean Air Act and other federal air quality legislation. .
43. Compare the energy released on combustion of 1 gallon of ethanol and 1 gallon of gasoline.
Assume gasoline is pure octane (C8H18). Explain the difference.
Answer:
Figure 4.16 gives the energy released per gram for the combustion of several fuels.
Assuming the densities of octane and ethanol are similar (a good assumption) one gallon of
gasoline releases more energy than one gallon of ethanol. (47.8 kJ/mole of gasoline vs. 29.7
kJ/mole of ethanol) This makes sense, because ethanol is an oxygenated fuel, that is, it
contains oxygen and thus is already “partially burned.”
44. Your neighbor is shopping for a new family vehicle. The salesperson identified a van of
interest as a flexible fuel vehicle (FFV).
a. Explain what is meant by FFV to your neighbor.
b. What is E85 fuel?
c. Would your neighbor and his family be particularly interested in using E85 fuel depending
on what region of the country they live?
Answer:
a. Flexible Fuel Vehicles (FFVs) can be fueled with conventional gasoline, E85, or a
combination of the two within the same tank.
b. E85 is a blend of 85 percent ethanol and 15 percent gasoline.
c. The availability of E85 is the key factor here. For example, at present, E85 is most
available in upper Midwest states: Minnesota, Illinois, Iowa, Indiana, and Wisconsin.
PAGE 4-18
45. The concept of entropy and disorder is used in games like poker. Describe how the rank of
hands (from a simple high card to a royal flush) is related to entropy and probability.
Answer:
A royal flush is an ace, king, queen, jack, and ten of the same suit. It is a highly improbable
hand in poker (1 in about 650,000 five-card hands). It exhibits a higher degree of order (low
entropy) and is more highly valued than a simple high card hand (a higher degree of
entropy). The hand with the least entropy wins!
46. Bond energies such as those in Table 4.4 are sometimes found by “working backward” from
heats of reaction. A reaction is carried out, and the heat absorbed or evolved is measured.
From this value and known bond energies, other bond energies can be calculated. For
example, the energy change associated with the combustion of formaldehyde (H2CO) is 465
kJ.
Use this information and the values found in Table 4.4 to calculate the energy of the C=O
double bond in formaldehyde. Compare your answer with the C=O bond energy in CO2 and
speculate on why there is a difference.
Answer:
Let x represent the C=O bond energy in H2CO.
Bonds broken in the reactants:
2 mol C–H single bonds = 2(416 kJ) =
1 mol C=O double bonds = 1(x kJ)
=
Total energy absorbed by breaking bonds
832 kJ
x kJ
=
(832 + x) kJ
Bonds formed in the products
2 mol O–H single bonds = 2(467 kJ)
1 mol C=O double bonds = 1(803 kJ)
934 kJ
803 kJ
=
=
Net energy change: (+832 + x kJ) – (1737 kJ) = 465 kJ
Rearranging the equation: x kJ = 465 + 1737 – 832 kJ
x = 440 kJ
This value is much less than the bond energy for C=O double bonds in carbon dioxide
reported in Table 4.4. The C=O double bonds in carbon dioxide are much stronger than the
C=O double bond in formaldehyde.
Exploring the Extensions
PAGE 4-19
47. Revisit the Six Principles of Green Chemistry found on the inside of the front cover.
Which of these are met by the synthesis by Suppes of propylene glycol from glycerol? Hint:
see equation 4.15).
Answer:
Principle 1 (“It is better to prevent waste than to treat or clean up waste after it is formed.”) is met
because the original waste product is converted into a useful material. Principle 3 (“It is
better to use and generate substances that are not toxic.”) is also met because non-toxic propylene
glycol is made from toxic ethylene gycol. Principles 4 and 5 (“It is better to use less energy”
and “It is better to use renewable materials.”) also are met, because the starting materials in the
production of biodiesel are renewable, and because the propylene glycol is made from a byproduct, less energy is required.
48. Another claim in the Scientific American article by Lovins referenced in Section 4.11 was
that replacing an incandescent bulb (75 W) with a compact fluorescent bulb (18 W) would
save about 75% in the cost of electricity. Electricity is generally priced per kilowatt-hour
(kWh). Using the price of electricity where you live, calculate how much money you would
save over the life of one compact fluorescent bulb (about 10,000 hr).
Answer:
Answers will vary depending on the current cost of electricity. However, the “75% less
energy” claim can be easily validated.
Over the lifetime of a compact fluorescent, 180 kWh of electricity are used:
10,000 hours  0.018 kW = 180 kWh
Over the same amount of time, a standard light bulb will use 750 kWh of electricity,
10,000 hours  0.075 kW = 750 kWh
Over 10,000 hours, the compact fluorescent uses 24% of the electricity of a standard bulb.
180/750 = 24%
49. Section 4.7 states that RFGs burn more cleanly by producing less carbon monoxide than
nonoxygenated fuels. At the molecular level, what evidence supports this statement?
Answer:
RFGs are oxygenates. This means that these fuels contain oxygen in addition to carbon and
hydrogen. With more oxygen present in the fuel itself, it is more likely that the fuel will burn
completely to produce carbon dioxide. Carbon monoxide production should be minimized.
50. Another type of catalyst used in the combustion of fossil fuels is the catalytic converter that
was discussed in Chapter 1. One of the reactions that these catalysts speed up is the
conversion of NO(g) to N2(g) and O2(g).
a. Draw a diagram of the energy of this reaction similar to the one shown in Figure 4.20.
b. Why is this such an important reaction? Hint: See Sections 1.9 and 1.11.
PAGE 4-20
Answer:
a. The sketch shows that the catalyzed pathway requires less activation energy than the
uncatalyzed pathway.
b. In Chapter 1, catalysts were discussed in connection with reducing NO from automobile
exhaust. Nitrogen oxide can react with oxygen to form NO2, a criteria pollutant. NO is also
involved in forming ozone in the troposphere and contributes to acid rain. To reduce
pollution, it is important to reduce NO emissions.
51. Chemical explosions are very exothermic reactions. Describe the relative bond strengths in
the reactants and products that would make for a good explosion.
Answer:
Consider a natural gas (methane) explosion:
CH4 + 2 O2  CO2 + 2 H2O
The bond energies involved are: C–H single bonds, 416 kJ/mole; O=O double bonds, 498
kJ/mole; H–O single bonds, 467 kJ/mole; C=O double bonds, 803 kJ/mole. The bond
energies of the products are larger than those of the reactants. This will lead to a large
negative net energy change indicating an exothermic reaction.
52. Because the United States has large natural gas reserves, there is significant interest in
developing uses of this fuel. List two advantages and two disadvantages of using natural gas
to fuel vehicles.
Answer:
The advantages of using natural gas to fuel vehicles include: (1) the lower amounts of
pollutants released into the atmosphere such as VOCs, CO, NOx, and SOx, (2) it burns more
cleanly and completely, and (3) natural gas releases more energy per gram than gasoline.
Disadvantages include (1) the difficulty in transporting and dispensing a gas (compared to a
liquid) as well as the lack of “natural gas stations,” (2) like petroleum, natural gas is a nonrenewable fossil fuel, and (3) the combustion of natural gas produces carbon dioxide, a
PAGE 4-21
greenhouse gas.
53.
You may have seen some General Motors advertisements using the slogan “Live
Green by Going Yellow” for their FlexFuel vehicles that can use E85 gasoline. To
what do the colors in this slogan refer?
Answer:
Living “green” refers to living in an environmentally conscious way. GM is encouraging
people to use E85 gasoline to reduce consumption of fossil fuels. The ethanol used in E-85
gasoline in the U.S. is derived from corn, which is yellow.
54.
China’s large population has increased energy consumption as the standard of
living increases.
a. Report on China’s increasing number of automobiles over the last 10 years.
b. What evidence suggests that the increase in the number of vehicles has affected air
quality? What interventions, if any, does the Chinese government have underway?
Answer:
a. The last decade saw an explosion of automobiles in China. In 2009, China surpassed the
United States as the world’s largest automobile market, with over 13 million vehicles sold
that year alone. A large fraction of the cars are located in the large cities, especially Beijing,
whose car population exceeds 4 million; 2000 additional vehicles are being added every day.
b. Automobiles created pollution problems because of their sheer number, poor road
infrastructure, old vehicle technology, and no emission standards before the year 2000.
Health effects and poor crop yields are being attributed to the pollution created by this drastic
increase in automobile use. The Chinese government has already begun taken steps to try to
reduce the pollution by requiring new emission standards similar to those in Europe for many
large metropolitan areas. In 2009, the government announced a 1.5 billion dollar grant
program to help the Chinese auto manufacturers develop new electric car technology.
55. Quality of life and energy consumption are related as shown in Figure 4.26. What ethical
considerations (if any) about their lifestyle do citizens of a country having a per capita
consumption of 8000 kgoe have to the rest of the world?
Answer:
Possible answers include concern over the equal distribution of natural resources, and the
unequal share of environmental damage throughout the world. Other issues involve global
sustainability; the planet cannot currently support everyone living a lifestyle that includes
consuming 8000 kgoe. There are also issues of consumption excesses. From the graph, the
quality of life, at least as measured by the HEI, does not improve significantly above 3000
kgoe. With a limited amount of energy available, when the excess use by some means the
others will have less.
56. What are the advantages and disadvantages of replacing gasoline with renewable fuels such
as ethanol? Indicate your personal position on the issue and state your reasoning.
PAGE 4-22
Answer:
Ethanol derived from plant matter is a renewable energy source. Ethanol burns more cleanly
than gasoline. However, widespread use of ethanol as automobile fuel could divert millions
of acres of cropland from producing food to producing fuel. Cars using ethanol are less fuel
efficient than when using gasoline, and ethanol use may cause engine problems in some
vehicles. Personal positions, of course, depend on your values (and to a certain extent where
you live).
57.
According to the EPA, driving a car is “a typical citizen’s most polluting daily
activity.”
a. Do you agree? Explain.
b. What pollutants do cars emit? Hint: Information on automobile emissions provided by the
EPA (together with the information in this text) can help you fully answer this question.
c. RFGs play a role in reducing emissions. Where in the country are RFGs required? Check
the current list published on the web by the EPA.
d. Explain which emissions RFGs are supposed to lower.
Answer:
a. To answer this question, one must assess all of the polluting activities of a typical citizen
and then compare the pollution of these activities to that of driving. Polluting activities
include all those that require energy, including heating, cooling, using appliances, cooking,
and transportation. If the typical citizen drives an automobile, this is likely to be the most
polluting activity. In contrast, if the typical citizen uses public transportation or some other
less polluting means (walking, cycling), heating or cooling may be the activity that generates
the highest level of pollution.
b. Cars emit the criteria pollutants CO, particulate matter, and NO via their exhaust pipes.
They also emit CO2, a greenhouse gas.
c. Areas of the United States where RFGs are required include most of the highly populated
urban centers on the eastern seaboard, Los Angeles and much of the California’s central
valley, Dallas-Fort Worth, Houston, Chicago, and St. Louis.
d. RFGs contain oxygenates that are intended to reduce CO emissions. They are also lower
in VOCs, which helps to reduce ozone formation.
58.
Research the Three Gorges Dam in China. Investigate some of the major issues
concerning this dam. Present your findings in a format of your choice.
Answer:
This major project has both drawbacks and benefits. The benefits include hydroelectric
power and flood control on the Chang Jiang (Yangzi) River. The drawbacks include the loss
of the fertile land and the 1 million plus people who had to be relocated.
Environmental issues also are being debated. Some people predict that industrial waste will
accumulate in the reservoir formed behind the dam. Others say that the hydroelectric power
will take the place of millions of tons of coal that would have been mined otherwise, thereby
PAGE 4-23
saving the country from the environmental damage caused by mining. Archaeologists are
concerned about the loss of ancient sites.
59. C. P. Snow, a noted scientist and author, wrote an influential book called The Two Cultures,
in which he stated: “The question, ‘Do you know the second law of thermodynamics?’ is the
cultural equivalent of ‘Have you read a work of Shakespeare’s?’” How do you react to this
comparison? Discuss his remark in light of your own educational experiences.
Answer:
At least one of the authors of this text does not like this comparison. She feels that neither of
these questions necessarily means anything in and of itself. A person might be able to recite
the second law of thermodynamics, and yet not understand its universal significance.
Similarly, a person might have read a play by William Shakespeare and yet not be able to
gain any perspective on human relations. A more meaningful comparison across the Two
Cultures could require a higher level of engagement. For example, consider wanting to
communicate something about an important societal topic – say global climate change in
some part of the planet. One way to do this might be to draft an environmental statement.
Another could be to create some form of art to convey the complexities of the climate.
PAGE 4-24