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CHAPTER 10
Energy
1.
energy
2.
Potential energy is energy due to position or composition. A stone at the top of a hill possesses
potential energy since the stone may eventually roll down the hill. A gallon of gasoline possesses
potential energy since heat will be released when the gasoline is burned.
3.
KE =
4.
The total energy of the universe is constant. Energy cannot be created or destroyed, but can only
be converted from one form to another.
5.
A state function is a property of a system which changes independently of the pathway taken to
make the change in the system. For example, the actual vertical distance between corresponding
points on the 2nd floor and on the 3rd floor in your dormitory is a property which does not change.
You might walk a different distance depending on whether you take the stairs or the elevator, but
the actual distance between the floors is a fixed distance.
6.
Ball A initially possesses potential energy by virtue of its position at the top of the hill. As Ball A
rolls down the hill, its potential energy is converted to kinetic energy and frictional (heat) energy.
When Ball A reaches the bottom of the hill and hits Ball B, it transfers its kinetic energy to Ball
B. Ball A then has only the potential energy corresponding to its new position.
7.
Temperature is a measure of the random motions of the particles in a substance, that is,
temperature is a measure of the average kinetic energies of the particles. Heat is the energy that
flows because of a temperature difference.
8.
The hot tea is at a higher temperature, which means the particles in the hot tea have higher
average kinetic energies. When the tea spills on the skin, energy flows from the hot tea to the
skin, until the tea and skin are at the same temperature. This sudden inflow of energy causes the
burn.
9.
The thermal energy of an object represents the random motions of the particles of matter which
constitute the object.
10.
Temperature is the concept by which we express the thermal energy contained in a sample. We
cannot measure the motions of the particles/kinetic energy in a sample of matter directly. We
know, however, that if two objects are at different temperatures, the one with the higher
temperature has molecules that have higher average kinetic energies than the object at the lower
temperature.
1 2
mv
2
211
Chapter 10:
Energy
11.
The system is the part of the universe upon which we want to focus attention. In a chemical
reaction, the system represents the reactants and products of the chemical reaction. The
surroundings include everything else in the universe.
12.
When the chemical system evolves energy, the energy evolved from the reacting chemicals is
transferred to the surroundings.
13.
An exothermic reaction is one which releases energy: therefore the products will have a lower
potential energy than did the reactants.
14.
exactly equal to
15.
Thermodynamics is the study of energy and energy transfers. The first law of thermodynamics is
the same as the law of conservation of energy, which is usually worded as “the energy of the
universe is constant”.
16.
internal
17.
ΔE represents the change in internal energy of the system (the internal energy represents the total
of the kinetic and potential energies of all the particles in the system); q represents the heat
energy that flows during a process; w represents the work done by, or upon, a system during a
process.
18.
losing
19.
positive
20.
gaining
21.
The calorie represents the amount of energy required to warm one gram of water by one Celsius
degree. The “Calorie” or “nutritional” calorie represents 1000 calories (one kilocalorie). The
Joule is the SI unit of energy and 1 calorie is equivalent to 4.184 Joules. The Joule has a specific
SI definition as the energy required to exert a force of 1 Newton over a distance of 1 meter, but
we tend to use the experimental definition above in terms of heating water.
22.
a.
1J
4.184 cal
b.
4.184 cal
1J
c.
1 kcal
1000 cal
d.
1000 J
1 kJ
23.
3 × 8.40 kJ = 25.2 kJ for three times the increase in temperature (a temperature increase of 15° as
opposed to 5°).
24.
6540 J = 6.54 kJ for 10 times more water
212
Chapter 10:
25.
26.
27.
28.
29.
4.184 kJ
= 315 kJ = 3.15 × 103 J
1 kcal
a.
75.2 kcal ×
b.
75.2 cal ×
c.
1.41 × 103 cal ×
d.
1.41 kcal ×
a.
8254 cal ×
1 kcal
= 8.254 kcal
1000 cal
b.
41.5 cal ×
1 kcal
= 0.0415 kcal
1000 cal
c.
8.231 × 103 ×
1 kcal
= 8.231 kcal
1000 cal
d.
752,900 cal ×
1 kcal
= 752.9 kcal
1000 cal
a.
652.1 kJ ×
b.
1.00 kJ ×
c.
4.184 kJ ×
d.
4.351 × 103 kJ ×
a.
243,000 J ×
b.
4.184 J ×
1 kJ
= 0.004184 kJ
1000 J
c.
0.251 J ×
1 kJ
= 0.000251 kJ
1000 J
d.
450.3 J ×
1 kJ
= 0.4503 kJ
1000 J
a.
625.2 cal ×
b.
82.41 kJ ×
4.184 J
= 315 J = 0.315 kJ
1 cal
4.184 J
= 5.90 × 103 J = 5.90 kJ
1 cal
4.184 kJ
= 5.90 kJ = 5.90 × 103 J
1 kcal
1 kcal
= 159.9 kcal
4.184 kJ
1 kcal
= 0.239 kcal
4.184 kJ
1 kcal
= 1.000 kcal
4.184 kJ
1 kcal
= 1040. kcal
4.184 kJ
1 kJ
= 243 kJ to 3 significant figures
1000 J
4.184 J 1 kJ
= 2.616 kJ
×
1 cal
1000 J
1000 J
= 8.241 × 104 J
1 kJ
213
Energy
Chapter 10:
30.
31.
Energy
1000 cal 4.184 J
×
= 2.201 × 105 J
1 kcal
1 cal
c.
52.61 kcal ×
d.
124.2 kJ ×
a.
91.74 kcal ×
b.
1.781 kJ ×
c.
4.318 × 103 J ×
d.
9.173 × 104 cal ×
1 kcal
4.184 kJ
= 29.68 kcal
1000 cal
= 9.174 × 104 cal
1 kcal
1 kcal
1000 cal
×
= 425.7 cal
4.184 kJ
1 kcal
1 cal
1 kcal
= 1.032 kcal
×
4.184 J 1000 cal
4.184 J 1 kJ
×
= 383.8 kJ
1 cal
1000 J
Q = s × m × ΔT
69,500 J = s × (1012 g) × (11.4°C)
s = 6.02 J/g °C
32.
Q = s × m × ΔT
Specific heat capacity of aluminum is 0.89 J/g°C from Table 10.1.
Q = (0.89 J/g°C) × (42.7 g) × (15.2°C) = 5.8 × 102 J (only two significant figures are justified).
33.
Q = s × m × ΔT
ΔT =
Q
s × m
Specific heat capacity of silver is 0.24 J/g°C.
ΔT =
34.
125 J
= 18°C (only two significant figures are justified)
0.24 g J°C × 29.3 g
Q = s × m × ΔT
Specific heat capacity of mercury is 0.14 J/g°C
100. J = (0.14 J/g°C) × 25 g × ΔT
ΔT = 28.6°C = 29°C
35.
Q = s × m × ΔT
Specific heat capacity of gold is 0.13 J/g°C
Q = (0.13 J/g°C) × (55.5 g) × (25°C) = 180 J = 1.8 × 102 J
214
Chapter 10:
36.
The reaction is exothermic: the chemical reaction liberates heat energy and warms up the
beverage.
37.
Q = s × m × ΔT
Energy
specific heat capacity of water = 4.184 J/g°C
Let’s assume that the density of water is exactly 1.00 g/mL during the measurement.
Q = 4.184 J/g°C × 1000 g × 15°C = 6.3 × 104 J = 63 kJ (~60 kJ to one significant figure)
38.
Q = s × m × ΔT
specific heat capacity of water = 4.184 J/g°C
Q = 4.184 J/g°C × 100.0 g × 35°C = 1.46 × 104 J (~15 kJ to two significant figures)
39.
The enthalpy change is the heat that flows on a molar basis for a reaction at constant pressure.
40.
A calorimeter is an insulated device in which reactions are performed and temperature changes
measured, enabling the calculation of heat flows. See Figure 10.6
41.
a.
The balanced equation as written represents the formation of two moles of HF. The
−542 kJ
enthalpy change per mole of HF will be
= –271 kJ/mol
2 mol
b.
The sign of the enthalpy of reaction is negative: therefore the reaction releases heat
energy and is exothermic.
c.
The conservation of energy principle requires that the heat of a reverse process be the
same in magnitude but opposite in sign. The enthalpy of reaction for the given equation is
–542 kJ. Therefore the enthalpy change of the reverse reaction will be +542 kJ.
a.
molar mass of S = 32.07 g
42.
1.00 g S ×
1 mol S
= 0.0312 mol S
32.07 g S
From the balanced chemical equation, combustion of 0.0312 mol S would produce
0.0312 mol SO2
0.0312 mol ×
b.
From the balanced chemical equation, combustion of 0.0312 mol S would produce
0.0312 mol SO2
0.501 mol ×
c.
43.
−296 kJ
= –9.23 kJ
mol
−296 kJ
= –148 kJ
mol
The enthalpy change would be the same in magnitude, but opposite in sign = +296
kJ/mol
The text indicates that 55 kJ of energy is released per gram of methane during the combustion.
Methane’s molar mass is 16.04 g.
215
Chapter 10:
–
44.
Energy
55 kJ
16.04 g
×
= 8.8 × 102 kJ/mol (to two significant figures)
1 gram
1 mol
a.
molar mass of ethanol = 46.07 g; we will assume that 1360 has only 3 significant figures.
–
1360 kJ
1 mol
×
= –29.5 kJ/g
1 mol
46.07 g
b.
Since energy is released by the combustion, ΔH is negative: ΔH = –1360 kJ
c.
In the reaction as written, three moles of water vapor are produced when one mole of
ethanol reacts.
1 mol C 2 H 5 OH
1360 kJ
×
= 453 kJ/mol H2O
1 mol C 2 H 5 OH
3 mol H 2 O
45.
46.
47.
If the second equation is reversed and then added to the first equation, the desired equation can be
generated:
X(g) + Y(g) → XY(g)
ΔH = a kJ (as given)
XZ(g) → X(g) + Z(g)
ΔH = –b kJ (the equation was reversed)
Y(g) + XZ(g) → XY(g) + Z(g)
ΔH = a + [–b] kJ or (a – b) kJ
The desired equation 2C(s) + O2(g) → 2CO(g) can be generated by taking twice the first
equation and adding it to the reverse of the second equation:
2 × [C(s) + O2(g) → CO2(g)]
ΔH = 2 × –393 kJ = –786 kJ (equation doubled)
2CO2(g) → 2CO(g) + O2(g)
ΔH = –(–566 kJ) = +566 kJ (equation reversed)
2C(s) + O2(g) → 2CO(g)
ΔH = (–786) + (+566) = –220 kJ
The desired equation S(s) + O2(g) → SO2(g) can be generated by reversing the second equation
and dividing the equation by 2, and then adding this to the first equation:
S(s) +
48.
49.
3
2
O2(g) → SO3(g)
ΔH = –395.2 kJ (as given)
2SO3(g) → 2SO2(g) + O2(g)
ΔH = –(–198.2 kJ)/2 = +99 kJ
S(s) + O2(g) → SO2(g)
ΔH = (–395.2 kJ) + (+99 kJ) = –296.1 kJ
The desired equation can be generated as follows:
NO(g) + O3(g) → NO2(g) + O2(g)
ΔH = –199 kJ
½ × [3O2(g) → 2O3(g)]
ΔH = ½ × –(–427 kJ) = +213.5 kJ
½ × [2O(g) → O2(g)]
ΔH = ½ × –(+495 kJ) = –247.5 kJ
NO(g) + O(g) → NO2(g)
ΔH = (–199) + (+213.5) + (–247.5) = –233 kJ
The energy is converted to the energy of motion of the car, and to friction as the car’s tires
interact with the road. Once the potential energy has been dispersed, it cannot be reused.
216
Chapter 10:
Energy
50.
Once everything in the universe is at the same temperature, no further thermodynamic work can
be done. Even though the total energy of the universe will be the same, the energy will have been
dispersed evenly making it effectively useless.
51.
Petroleum is especially useful because it is a concentrated, easy to transport and use, source of
energy.
52.
Concentrated sources of energy, such as petroleum, are being used so as to disperse the energy
they contain, making it unavailable for further use.
53.
These sources of energy originally came from living plants and animals, which used their
metabolic processes to store energy.
54.
Petroleum consists mainly of hydrocarbons, which are molecules containing chains of carbon
atoms with hydrogen atoms attached to the chains. The fractions are based on the number of
carbon atoms in the chains: for example, gasoline is a mixture of hydrocarbons with 5–10 carbon
atoms in the chains, whereas asphalt is a mixture of hydrocarbons with 25 or more carbon atoms
in the chains. Different fractions have different physical properties and uses, but all can be
combusted to produce energy. See Table 10.3
55.
Natural gas consists primarily of methane, with small amounts of ethane, propane and butane. It
is generally found in association with petroleum deposits.
56.
Tetraethyl lead was used as an additive for gasoline to promote smoother running of engines. It is
no longer widely used because of concerns about the lead being released to the environment as
the leaded gasoline is burned.
57.
Coal matures through four stages: lignite, sub-bituminous, bituminous, and anthracite. The four
types of coal differ in the ratio of carbon to the other elements. Anthracite has the highest fraction
of carbon in it, and when burned, releases a larger amount of heat for a given mass.
58.
The greenhouse effect is a warming effect due to the presence of gases in the atmosphere which
absorb infrared radiation that has reached the earth from the sun, and do not allow it to pass back
into space. A limited greenhouse effect is desirable because it moderates the temperature changes
in the atmosphere that would otherwise be more drastic between daytime when the sun is shining
and nighttime. Having too high a concentration of greenhouse gases, however, will elevate the
temperature of the earth too much, affecting climate, crops, the polar ice caps, temperature of the
oceans, and so on. Carbon dioxide produced by combustion reactions is our greatest concern as a
greenhouse gas.
59.
A driving force is some factor that tends to make a process occur.
60.
If a proposed reaction involves either or both of those phenomenon, the reaction will tend to be
favorable.
61.
an increase in entropy
62.
Formation of a solid precipitate represents a concentration of matter.
63.
Entropy is a measure of the randomness or disorder in a system. The entropy of the universe
increases because the natural tendency is for things to become more disordered.
217
Chapter 10:
Energy
64.
The molecules in liquid water are moving around freely, and are therefore more “disordered” than
when the molecules are held rigidly in a solid lattice in ice. The entropy increases during melting.
65.
As the steam is cooled from 150°C to 100°C, the molecules of vapor gradually slow down as they
lose kinetic energy. At 100°C, the steam condenses into liquid water, and the temperature remains
at 100°C until all the steam has condensed. As the liquid water cools, the molecules in the liquid
move more and more slowly as they lose kinetic energy. At O°C, the liquid water freezes.
66.
a.
462.4 kJ ×
1 kcal
= 110.5 kcal
4.184 kJ
b.
18.28 kJ ×
1 kcal
= 4.369 kcal
4.184 kJ
c.
1.014 kJ ×
1 kcal
= 0.2424 kcal
4.184 kJ
d.
190.5 kJ ×
1 kcal
= 45.53 kcal
4.184 kJ
a.
85.21 cal ×
4.184 J
= 356.5 J
1 cal
b.
672.1 J ×
c.
8.921 kJ ×
d.
556.3 cal ×
67.
68.
1000 J
= 8921 J
1 kJ
4.184 J
1 kJ
×
= 2.328 kJ
1 cal
1000 J
Temperature increase = 75.0 – 22.3 = 52.7°C
145 g × 4.184
69.
1 cal
= 160.6 cal
4.184 J
J
1 cal
× 52.7°C ×
= 7641.5 cal = 7.65 kcal
g °C
4.184 J
Specific heat of silver = 0.24 J/g°C; 1.25 kJ = 1250 J
Temperature increase = 15.2 – 12.0 = 3.2°C
Q = s × m × ΔT
1250 J = (0.24 J/g°C) × (mass of silver) × (3.2°C)
mass of silver = 1627 g = 1.6 × 103 g
70.
From Table 10.1, the specific heat capacity of iron is 0.45 J/g°C.
Q = s × m × ΔT
Q = 0.45
J
× 25.1 g × 17.5°C = 197.7 J = 2.0 ×102 J (two significant figures)
g°C
218
Chapter 10:
Energy
J
1 cal
cal
= 0.31
×
4.184 J
g°C
g°C
71.
0.13
72.
2.5 kg water = 2500 g
Temperature change = 55.0 – 18.5 = 36.5°C
Q = s × m × ΔT
Q = 4.184 J/g°C × 2500 g × 36.5°C = 3.8 × 105 J
73.
For a given mass of substance, the substance with the smallest specific heat capacity (gold,
0.13J/g°C) will undergo the largest increase in temperature. Conversely, the substance with the
largest specific heat capacity (water, 4.184 J/g°C) will undergo the smallest increase in
temperature.
74.
Let Tf represent the final temperature reached by the system.
For the hot water, heat lost = 50.0 g × 4.184 J/g°C × (100. – Tf°C)
For the cold water, heat gained = 50.0 g × 4.184 J/g°C × (Tf – 25°C)
The heat lost by the hot water must equal the heat gained by the cold water; therefore
50.0 g × 4.184 J/g°C × (100. – Tf°C) = 50.0 g × 4.184 J/g°C × (Tf – 25°C)
Solving this equation for Tf gives Tf = 62.5°C = 63°C
75.
Let Tf be the final temperature reached.
Heat gained by water = 75 g × 4.184 J/g°C × (Tf – 20°C)
Heat lost by iron = 25.0 g × 0.45 J/g°C × (85 – Tf°C)
The heat lost by the iron must equal the heat gained by the water.
75 g × 4.184 J/g°C × (Tf – 20°C) = 25.0 g × 0.45 J/g°C × (85 – Tf°C)
Solving for Tf gives Tf = 22.3°C = 22°C
76.
9.0 J (It requires twice as much heat to warm a sample of twice the mass over the same
temperature interval.)
77.
For any substance, Q = s × m × ΔT . The basic calculation for each of the substances is the
same: Heat required = 150. g × (specific heat capacity) × 11.2 °C
Substance
Specific Heat Capacity
Heat Required
water (l)
water (s)
water (g)
aluminum
iron
mercury
carbon
silver
gold
4.184 J/g °C
2.03 J/g °C
2.0 J/g °C
0.89 J/g °C
0.45 J/g °C
0.14 J/g °C
0.71 J/g °C
0.24 J/g °C
0.13 J/g °C
7.03 × 103 J
3.41 × 103 J
3.4 × 103 J
1.5 × 103 J
7.6 × 102 J
2.4 × 102 J
1.2 × 103 J
4.0 × 102 J
2.2 × 102 J
219
Chapter 10:
78.
79.
80.
81.
Energy
Because, for any substance, Q = s × m × ΔT , we can solve this equation for the temperature
change, ΔT. The results are tabulated:
Substance
Specific Heat Capacity
Temperature Change
water (l)
water (s)
water (g)
aluminum
iron
mercury
carbon
silver
gold
4.184 J/g °C
2.03 J/g °C
2.0 J/g °C
0.89 J/g °C
0.45 J/g °C
0.14 J/g °C
0.71 J/g °C
0.24 J/g °C
0.13 J/g °C
23.9 °C
49.3 °C
50. °C
1.1 × 102 °C
2.2 × 102 °C
7.1 × 102 °C
1.4 × 102 °C
4.2 × 102 °C
7.7 × 102 °C
a.
ΔE= q + w = –47 kJ + 88 kJ = 41kJ
b.
ΔE = 82 + 47 = 129 kJ
c.
ΔE = 47 + 0 = 47 kJ
d.
When the surroundings deliver work to the system, w > O. This is the case for a and b.
a.
The combustion of gasoline releases heat, so this is an exothermic process.
b.
H2O(g) → H2O(l); Heat is released when water vapor condenses, so this is an exothermic
process.
c.
To convert a solid to a gas, heat must be absorbed, so this is an endothermic process.
d.
Heat must be added (absorbed) in order to break a bond, so this is an endothermic
process.
4Fe(s) + 3O2(g) → 2Fe2O3(s) ΔH = –1652 kJ;
Note that 1652 kJ of heat are released when 4 mol Fe reacts with 3 mol O2 to produce 2 mol
Fe2O3.
−1652 kJ
= –1652 kJ
4 mol Fe
a.
4.00 mol Fe ×
b.
1.00 mol Fe2O3 ×
c.
1.00 g Fe ×
d.
10.0 g Fe = 0.179 mol Fe; 2.00 g O2 = 0.0625 mol O2; O2 is the limiting reactant.
−1652 kJ
= –826 kJ
2 mol Fe 2 O3
1 mol Fe −1652 kJ
×
= –7.39 kJ
55.85 g Fe 4 mol Fe
0.0625 mol O2 ×
−1652 kJ
= –34.4 kJ heat released.
3 mol O 2
220
Chapter 10:
82.
Energy
Reversing the first equation and dividing by 6 we get
3
6
D →
or
1
2
3
6
ΔH = +403 kJ/6
A+B
D →
1
2
ΔH = +67.2 kJ
A+B
Dividing the second equation by 2 we get
1
2
E+F →
1
2
ΔH = –105.2 kJ/2 = –52.6 kJ
A
Dividing the third equation by 2 we get
1
2
C →
1
2
E+
3
2
D
ΔH = +64.8 kJ/2 = +32.4 kJ
Adding these equations together we get
1
2
83.
C + F → A + B + D ΔH = 47.0 kJ
During exercise, the body generates about 5500 kJ/hr, or about 11,000 kJ in 2 hours. Assuming
all of this heat is lost through evaporation, we can calculate the volume of the perspiration. Water
has a heat of vaporization of 40.7 kJ/mol. So,
11,000 kJ ×
1 mol H 2 O
= 270 mol H2O
40.7 kJ
270 mol H2O ×
18.02 g H 2 O
= 4900 g H2O
1 mol H 2 O
Assuming a density of 1 g/mL for water, 4900 g of H2O would occupy a volume of 4900 mL or
4.9 L.
84.
400 kcal is burned per hour walking 4.0 mph, then the total amount of heat burned while walking
at 4.0 mph will be given by
Q = 400 kcal/hr × t
where t is the amount of time spent walking (in hours). One gram of fat is consumed for every 7.7
kcal of heat. Therefore, one pound of fat requires
7.7 kcal/g × 454 g/lb = 3500 kcal/lb
Since we want to lose one pound, Q = 3500 kcal and
t=
Q
3500 kcal
=
= 8.75 hr = 9 hr
400 kcal/hr 400 kcal/hr
221