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Solutions to Chapter 7 Problems
Section 7.1. Energy and Change
7.1. A slice of cheese pizza typically contains 180 Calories of food energy. The human
heart requires about 1 J of energy for each beat.
(a) Assuming that when you metabolize the pizza, all of the food energy will be
used to keep your heart beating, how many heart beats can this slice of pizza
sustain? About how many minutes would this keep your heart going? Hint : 1
Calorie = 1 kcal.
(b) Assume the efficiency of use of this food energy is about the same as that
calculated for capturing the energy from glucose oxidation, Section 7.9. What are
your answers for part (a) in this case? Explain your reasoning.
Answer to 7.1:
180 Cal×1 κχαλ
1 Χαλ×
4184 κϑ
1 κχαλ×
1000 ϑ
1 κϑ×
1 ηεαρτβεατ
1 ϑ=8×10
8 ηεαρτβεα
7.2. Pasta is largely starch and a deep-fried potato chip is mostly starch and fat.
(a) What do your observations from Investigate This 7.1 tell you about the
energy value of these foods? Do they combine exothermically with oxygen? What
is the evidence and reasoning for your response?
(b) Starch is a polymer of glucose. Is this composition consistent with your
answers in part (a) and other data from Investigate This 7.1? Explain your
reasoning.
Answer to 7.2:
(a) Once they have been ignited both the pasta and potato chip continue to
burn on their own. They release energy in the form of light and enough
thermal energy to keep their temperatures high enough to continue
burning. You might find that a potato chip burns somewhat better than
pasta, because the fat is a better fuel than starch. This is reflected in their
nutritional Calorie values as well. A gram of starch provides about 4
Calories (17 kJ) and a gram of fat about 9 Calories (38 kJ).
(b) The oxidation of glucose has been used as an example throughout the chapter.
You know that the oxidation is quite exothermic, which is shown by the burning
pasta, potato chip, and marshmallow (mainly sugars including glucose). It doesn’t
matter that the glucose units are strung together as a polymer in starch or
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separated in a marshmallow -- each still oxidizes to produce a large output of
enthalpy.
7.3. Predict how the results of Investigate This 7.1 would differ if a baked, rather than
a fried, potato chip were used. What will happen if a raw potato slice is used?What do you predict will be observed if a chip using Olestra® is used? In each
case, explain the reasons for your prediction.
Answer to 7.3:
The baked chip will still burn, but it is predicted not to burn as vigorously as the
fried chip because there is less residual fat available as fuel on the baked chip.
The raw potato slice may not burn at all because it has a high water content. The
Olestra® chip will also burn quite well, for the Olestra is a long-chain polymeric
fatty acid. It does not produce heat value in the body because it cannot be
metabolized.
Section 7.2. Thermal Energy (Heat) and Mechanical Energy (Work)
7.4. Several different types of energy have been discussed in this chapter. These
include kinetic energy, mechanical energy (work), potential energy, and thermal
energy (heat). How are these terms related?
Answer to 7.4:
Potential energy and kinetic energy are the two types of energy.
Mechanical energy and thermal energy are both examples of kineticenergy. This is a visual representation of the relationship.
mecancaener ter7.5. Figure 7.2 illustrates a model for the movement of combustion gases.
(a) Draw a similar diagram for a tethered helium-filled balloon. Hint: Unlike
Figure 7.2, there will only be one type of particle, representing helium atoms.
(b) Will the helium atoms in the tethered balloon exhibit any directed motion?
Why or why not? What will happen to the motion of the helium atoms in the
balloon when the tether is released? Explain briefly.
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Answer to 7.5:
(a) Tethered Balloon
(b) The helium atoms in the balloon are in constant, random motion. They
do not exhibit any direction motion. The helium atoms in the balloon are
still in constant, random motion, keeping the balloon fully inflated. The
balloon itself will drift upward, buoyed by the air molecules around it.
7.6. Consider a drop of water in a waterfall. At the very top of the fall, the molecules
in the drop are moving randomly. While it is falling (pulled down by gravity), its
molecules have an overall downward motion. When it hits the pool at the bottom,
the molecular motion again becomes random.
(a) The water at the bottom of the waterfall is a little warmer than the water atthe top. What can you say about the change in internal energy of your drop of
water as it goes from the top of the waterfall to the bottom?
(b) Has work been done or has thermal energy been transferred to cause the
change in the internal energy of the water drop? Explain where the work and/or
thermal energy come from.
Answer to 7.6:
(a) Since the temperature of the water increases when it falls and hits the
pool, its internal energy must have increased.(b) Work has been done on the water to increase its internal energy (and, hence,
its temperature). The work is done on the water when its potential energy at the
top of the fall is converted to kinetic (directed) motion as it is falling and then this
directed motion is changed to random motion when it hits the pool at the bottom
and rapidly decelerates due to the force of the collision.
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7.7. When heat is added to H2O to do work, why does the vapor form of water
generate more work than either ice or liquid water?
Answer to 7.7:
Water molecules in ice are tightly ordered in the crystal lattice. These
molecules are held in place by hydrogen bonds that move only slightly
within this lattice. Liquid water molecules also hydrogen bond, but are
capable of breaking and remaking these relatively weak intermolecular
attractions as these water molecules move relative to one another. Keep in
mind that the molecules of liquid water move slowly compared to water
molecules in the vapor phase. There is more potential energy than kinetic
energy associated with these two phases of water. In the vapor phase,
where water molecules are on average, quite distant from their neighbors,
hydrogen bonding is negligible. All the heat energy added to the vapor
phase water is transformed into kinetic energy and from there into useful
mechanical work. On the other hand, heat energy added to liquid water is
used primarily to break hydrogen bonds and only secondarily to increase
the average kinetic energy of the molecules.
7.8. Why is it important to you, as a consumer, for automotive engineers to design
engines that convert the highest possible percentage of energy to work?
Answer to 7.8:
Economic and environmental factors are important to every consumer. Whatever
the price of gasoline, consumers want to obtain the maximum number of miles
possible per gallon of gasoline. This can only be done if the chemical energy from
the fuel is converted efficiently as possible to work, and not wasted as heat.
Efficient conversion of fuel energy to work also means release of lower
concentrations of carbon monoxide and hydrocarbons into the atmosphere.
Section 7.3. Thermal Energy (Heat) Transfer
7.9. For each case below, is thermal energy being transferred by radiation or by a
contact process of conduction and/or convection? Briefly justify your choice.
(a) A silver spoon at room temperature, when placed in a cup of hot water,
becomes too warm to touch.
(b) An apple pie is baked in an electric oven.
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(c) The water in an outdoor swimming pool cools from 25 °C to 20 °C as the
summer season changes into autumn.
(d) You sunbathe with your back exposed to the sun, until your back feels very
warm.
Answer to 7.9:
(a) The process of thermal energy transfer is by contact , using conduction.
The water molecules are moving more rapidly, on the average, than the
silver atoms in the spoon. Thermal energy is transferred from the warmer,
more energetic water molecules to the cooler, less energetic silver atoms
in the spoon, causing it to become uncomfortably warm.
(b) Thermal energy transfer is likely taking place through contact . Both
conduction and convection play a role in baking this apple pie. The
electric heating unit will transfer thermal energy to molecules of gases in
the air directly in contact with the unit; this is conduction. The warmer air
rises, allowing the denser, cooler air to circulate near the heating coil; this
sets up a convection current. (In ovens advertised as “convection ovens”,
there is at least one fan to help distribute the air evenly. ) The molecules of
hot gas that are in contact with the pie will conduct thermal energy into the
pie, eventually resulting in baking the pie.
(c) The water temperature is going down because thermal energy is being
lost by contact between warmer water molecules and the cooler air.
Conduction is involved directly at the water/air interface, but convection
currents in both water and air are likely the major factors in this heat
transfer process.
(d) Thermal energy is being transferred from the sun to your back by means of the
long wavelength infrared electromagnetic radiation. As the radiation is absorbed
by the skin on your back, the molecules in your skin are moving faster which
increases their temperature. Infrared radiation is accompanied by the more
energetic ultraviolet rays, so remember to use sunscreen to avoid skin damage.
7.10. At the end of a skating season, an indoor ice rink was closed. Without any outside
cooling, how will the roof and the ice rink floor reach a common temperature?
Will it be a slow or fast process? Explain your answer.
Answer to 7.10:
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Since the cooler air near the ice rink floor has a higher density, it will stay
stagnant for a relatively long time. The ice rink floor will remain cool until it is
warmed up by heat conduction from the roof or the outside through the walls.
Heat will be conducted by the interactions of warm and cool air molecules. This
process is not going to be fast. Convection is very inefficient in this situation.
7.11. After being immersed in cold water, the core body temperature of an individual
may be abnormally low. Design three methods, one using convection, the second
using conduction, and the third using radiation to transfer heat to warm up an
individual as quickly and safely as possible. Which method will be the most
efficient in this case?
Answer to 7.11:
The convection method might consist of a small chamber with a heated floor.
The problem with this heating method is that a border of stagnant, cold air may
surround the individual. Since the cooler air near the individual has a higher
density, it will stay stagnant for a relatively long time and prevent efficient
warming.. The radiation method may consist of a large chamber with secured,
open flame inside. In this method, thermal energy will be transferred to the victim
from the flame by infrared electromagnetic radiation,. It is not going to be safe
and efficient method because of the risks of burning the individual. Also, since
radiation travels in straight lines, not all areas of the individual will be warmed
equally. Wrapping the victim in an electric blanket will be the most efficient
process of heating. In this method, heat is transferred by conduction. The
individual will be heated quickly and uniformly.
Section 7.4. State Functions and Path Functions
7.12. Your monthly bank statement gives the opening balance for your account, details
the transactions that have occurred, and reports the ending balance. Which aspects
of this statement correspond to state functions and which to path-dependent
functions?
Answer to 7.12:
The difference between the ending and initial balance corresponds to a
state function.
∆ Balance = Balanceend − Balancestart
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All of the transactions correspond to path-dependent functions. They describe all
the different expenditures and deposits that are responsible for the overall change
in balance.
7.13.The quarterback starts a play on his own 35 yard line, but is pushed back by thedefense to his own 25 yard line. He then successfully completes a pass to the
opponent’s 40 yard line. What is the number of yards gained in the play? Is the
number of yards gained a state function or a path-dependent function?
Answer to 7.13:
The play has moved the team from their own 35 yard line, past midfield, and to
the opponent’s 40 yard line. The team has moved the ball 25 yards closer to the
end zone, another example of a state function. The path taken does not matter,
only the fact that this is a first-down play successfully completed.
7.14. Today’s most advanced fossil-fuel burning power plants producing electricity
operate at an efficiency of about 42%. Suggest some reasons why a higher
percentage of the energy from burning the fossil fuels is not converted to work.
Do you think it is theoretically possible to convert 100% of the energy into work?
Explain your reasoning.
Answer to 7.14:
A significant fraction of the heat energy is transferred to the water used to cool
the turbines, and there are many sources of friction in the turbines and generator.Even when the electricity has been generated, there will be losses over long-
distance power lines usually required to reach the power grid. Most students will
intuitively recognize that realize this seems unlikely and the introduction to this
chapter tells them it is not possible. Heat losses must present serious limitations,
given that the best engineers have only found a 42% efficiency rate possible. A
more complete answer to this question will be possible after studying Chapter 8,
Entropy, where the Second Law of Thermodynamics is studied in more detail.
7.15. Consider two different cases in which a fully charged battery becomes totally
discharged.
Case 1: The battery is used to provide power to a flashlight.
Case 2: The battery is used to provide power for a child’s toy car.
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(a) Will the change from the battery being fully charged to being totally
discharged be a state function or a path-dependent function in each case? Explain
your reasoning.
(b) Will there be any thermal energy transfer by radiation or by contact in either
use of the battery? Explain your reasoning?
Answer to 7.15:
(a) The change from the battery being fully charged battery to being
totally discharged battery is a state function in both cases. The path is
quite different, but the end result is exactly the same, a dead battery. State
functions do not depend on the path taken to achieve the present state.
(b) In Case 1, much of the energy lost by the battery will be released as radiant
energy, light. Batteries do get warm as they are being used, also indicating that
there is some thermal energy transfer. In Case 2, there will be some thermal
energy transfer because of friction between the toy car and the roadbed on which
the car runs. Much of the energy lost will be changed to mechanical work done in
moving the toy car.
7.16. Consider a C-172 airplane (small private plane), a helicopter, a parachute jumper,
and a hawk, all presently at 1892 feet above the Clemson, South Carolina airport,
which is at an altitude of 892 feet. Each plans a landing at the airport. Comment
on the state functions and path functions for each of these airborne objects.
Answer to 7.16:
The state functions will be identical, as each object will make an altitude change
of 1000 feet to land safely on the ground. The path functions will be quite
different, however. The airplane and helicopter are obliged to follow certain
traffic patterns, with designated altitudes and descent rates into the airport. The
parachute jumper will likely be following a spiral pattern to land safely. The bird
will take whatever path it wants to, with little regard to the airplane, helicopter, or
parachute jumper. Interestingly enough, the bird also lands into the wind, just as
the other airborne objects plan to do!
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Section 7.5. System and Surroundings
7.17. Which ending(s) for the following sentence is(are) correct? Explain your
reasoning in each case. The enthalpy change for a system open to the atmosphere
is(a) dependent on the identity of the reactants and products
(b) zero
(c) negative
(d) qP
(e) positive
Answer to 7.17: (a) and (d)
7.18. Give “real life” examples of the following:
(a) an open system(b) a closed system
(c) an isolated system
(d) an exothermic process
(e) an endothermic process
Answer to 7.18:
(a) An open system refers to part of the universe being studied that can
exchange matter and energy with its surroundings. Some open system
examples include an open aquarium, a cup of coffee, and hot springs. Ourbody is also an open system since we exchange both energy (in the form
of heat) and matter (food, O2, CO2, H2O) with the surroundings.
(b) In a closed system, only energy is exchanged. An example of such
system is a cup with hot coffee that is tightly closed so no vapors may
escape but the cup is not isolated so heat energy may be exchanged with
surroundings.
(c) An isolated system does not exchange matter or energy. An example of
a nearly isolated system might be a properly stoppered Thermos® bottle.A stopper prevents water vapor from escaping, while the vacuum
construction keeps heat from being lost to the surroundings.
(d) An exothermic process is a process that releases heat to its
surroundings. Reactions of sodium metal with water, burning of a match
or solidification of molten metals are examples exothermic processes.
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(e) An endothermic process is a process that absorbs heat into its surroundings.
Melting ice or evaporation of rubbing alcohol are examples of endothermic
processes.
7.19.A properly stoppered Thermos bottle containing coffee is nearly an isolatedsystem. The stopper prevents water vapor from escaping, while the vacuum
construction keeps heat from being lost to the surroundings. How would you turn
this isolated system into an open system? A closed system?
Answer to 7.19:
An open system can exchange matter and energy between system and
surroundings. Opening the stopper would change Thermos bottle into open
system in which matter (vapors) and heat can be transferred. A closed system can
exchange energy with its surroundings, but no matter. If, for example, the stopper
of Thermos bottle is pierced with a piece of heavy gauge copper wire, heat
would leave the system by conduction along the copper wire. Removing the
vacuum construction would also convert the isolated system of Thermos bottle
into an open system in which heat can be exchanged through the walls of the
bottle.
7.20. The earth is sometimes described as a closed system, particularly when
considering regional or global environmental problems. For example, acid rain
may fall at great distance from the original site of emissions responsible for theobserved effect. Ozone depletion near the South Pole is linked to the emission of
fluorocarbons from uses in industrial nations. While this view does convey a
sense of the many connections that affect environmental issues, is it correct to
think of the earth as a closed system from the standpoint of thermodynamics?
Why or why not?
Answer to 7.20:
The earth is an open system because matter and energy is exchanged on a regular
basis with the surroundings. For example, molecules escape the earth’s surfaceregularly, placing gases that are responsible for acid rain into the atmosphere.
Meteorites fall to the earth from outer space. Thermal energy is transferred from
the earth to the atmosphere, and various types of electromagnetic energy are
transferred to the earth’s surface.
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7.21. This diagram shows an
experiment in which a hot silver
bar is added to a beaker of water
at room temperature.
(a) If the object of the
experiment is to find the specific
heat of the silver, define the
system and the surroundings.
(b) If you already know the specific heat of the silver, the same experimental set
up might be used to determine if there is heat loss through the walls of the glass
beaker or from the thermometer itself. Define the system and surroundings in that
experiment.
(c) Does this experimental diagram illustrate an open, closed, or isolated system?
Give the reasoning behind your choice.
Answer to 7.21:
(a) The silver bar and the water form the collection of atoms and
molecules under consideration, the system. Everything else, including the
beaker, the thermometer, the air above the water, and everything else in
the universe are surroundings. There will be thermal energy transfer from
the warmer silver bar to the cooler water molecules until equilibrium is
reached. At that point, the heat released by the silver bar equals the heat
gained by the water. It is necessary to know the mass of the silver bar, the
mass of water present, and the specific heat of water.
(b) The silver bar, the water, the beaker, and the thermometer are all part
of the system in this experiment. Everything else forms the surroundings.
(c) This is an open system because matter and energy can be exchanged with the
surroundings. For example, water molecules can evaporate from or condense onto
the surface of the water. Heat may be exchanged between the system and the
surroundings. Therefore, this cannot be a closed system because matter can be
exchanged. It is not an isolated system as there is no attempt to control thermal
exchanges between the system and the surroundings.
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Section 7.6. Calorimetry and Introduction to Enthalpy
7.22. In a calorimetric measurement, why might it be important to know the heat
capacity of the calorimeter? The calorimeter heat capacity is the amount of
thermal energy the calorimeter (container, thermometer,etc
.) transfers per degreetemperature change.
Answer to 7.22:
To accurately measure the thermal energy change for reaction, we need to account
for heat that goes to heating (in the case of exothermic reaction) or cooling (in the
case of endothermic reaction) the calorimeter itself. The heat lost or gained to the
calorimeter must be experimentally determined for each calorimeter to assure the
accuracy of calorimetric measurements.
7.23. Choose the best response to complete the final sentence. Explain why you makethis choice and what is wrong with each of the others. The equation for
decomposition of gaseous ammonia, NH3(g), is:
NH3(g) → 1/2N2(g) + 3/2 H2(g) ∆ H = -45.9 kJ
This reaction equation and enthalpy change indicate that the formation of gaseous
ammonia
(a) evolves 45.9 kJ for each mole of ammonia formed.
(b) evolves 23 kJ for each mole of nitrogen used.
(c) absorbs 45.9 kJ for each mole of ammonia formed.(d) absorbs 23 kJ for each mole of nitrogen used.
(e) is an exothermic process.
Answer to 7.23: (c)
7.24. A 10.0 g sample of an unknown metal was heated to
100.0 °C and then added to 20.0 g of water at 23.0 °C in
an insulated calorimeter. At thermal equilibrium, the
temperature of the water in the calorimeter was 25.0 °C.
Which of the metals in this list is most likely to be the
unknown?
Answer to 7.24:
When the hot metal is placed in the cooler water, the heat lost by the metal
and the heat gained by the water sum to zero.
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Specific Heat Data
H2O(l) 4.184 J·g–1·°C–1
Au(s) 0.13 J·g–1·°C–1
Ag(s) 0.22 J·g–1·°C–1
Al(s) 0.90 J·g–1·°C–1
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heat lost + heat gained = 0
heat lost = –heat gained
mmetal × smetal × (Tf, metal–Ti, metal) = –m H2O × s H2O × (Tf, H2O–TI, H2O)
10 g × smetal × (25.0 °C – 100.0 °C) = –20 g water × 4.184 J
g ⋅°Χ × (25.0 °C – 23.0
°C)
specific heat metal = 0.22 J·g-1·°C-1 The metal is likely to be silver.
7.25. A student received 55.0 g of unknown metal from his laboratory instructor. To
identify this metal, he decided to calculate its specific heat. He heated the metal
sample to a temperature of 98.6 °C in a water bath. Then he transferred it slowly
to a calorimeter containing 100. mL of water at 24.6 °C. The maximum
temperature reached by the calorimeter system was 25.8 °C. His lab instructor,
who was watching the experiment, explained to the student that he hadintroduced a large amount of error and asked him to repeat the experiment with
the correct procedure. In the second trial, the temperature of the water bath was
98.4 °C and the initial temperature of water in the calorimeter was 24.3 °C. The
maximum temperature after the sample was immersed in the calorimeter was 26.5
°C. What error had the student made? Explain how you know.
Answer to 7.25:
The student transferred the hot material into the calorimeter too slowly. The heat
lost from metal was transferred to the air outside the calorimeter and it was not
equal to the heat gained by the water in the calorimeter. Therefore, the student
measured a lower final temperature of the water for the first trial.
7.26. A student mixed 100. mL of 0.5M HCl and 100. mL of 0.5M of NaOH in a
Styrofoam® cup calorimeter. The temperature of the solution increased from
19.0 °C to 22.2 °C. Is this process exothermic or endothermic? Assuming that the
calorimeter absorbs only a negligible quantity of heat, that the density of the
solution is 1.0 g·mL–1, and that its specific heat is 4.18 J·g–1·°C–1, calculate the
molar enthalpy change for the reaction:HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Answer to 7.26:
Total volume of solution is 200 mL . The mass of the solution is: (200
mL) x 1g/mL = 200 g
The temperature change is 22.2oC – 19.0oC = 3.2oC.
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The temperature increased. This means that thermal energy is released
from the system and the reaction is exothermic.
The heat gained by water is given by the product of the mass, specific
heat, and temperature change.
qp = (200 g) x (4.18 J/goC) x (3.2oC) = 2700 J =2.7 kJ
The heat gained by the system (reacting chemicals) is q= -2.7kJ.
To find the heat evolved per mole of reactants (enthalpy change) we
have to calculate the number of moles.
(0.1L) x (0.5 mol/L) = 0.050 mole
∆ H = -2.7kJ / 0.05 mole = -54kJ/mole
7.27. Instant cold packs, such as those used to treat athletic injuries, contain ammoniumnitrate and a separate pouch of water. When the pack is activated by squeezing to
break the water pouch, the ammonium nitrate dissolves in water. This is an
endothermic reaction and the change in enthalpy is 25.7 kJ·mol –1 of ammonium
nitrate dissolved.
(a) Why does this endothermic reaction produce a cold sensation on your skin?
(b) The cold pack contains 125 g of water to dissolve 50.0 g of ammonium
nitrate. What will be the final temperature of the activated cold pack, if the initial
room temperature is 25 °C? Assume that the specific heat of the solution is thesame as that for water, 4.184 J·g-1·°C-1.
(c) What other assumptions do you make in carrying out the calculation in part
(b)?
(d) The final temperature is about 7 °C. Does this result justify the assumptions
in parts (b) and (c)? If not, explain which assumption(s) might not be valid.
Answer to 7.27:
(a) The solution process is endothermic and is drawing heat from your
bruised skin to enable the ammonium nitrate to dissolve.(b) If the cold pack contains ample water to dissolve 50.0 g of ammonium
nitrate,
50.0g NH 4NO3 ×1 µολ ΝΗ 4ΝΟ380.0 γ ΝΗ 4 ΝΟ3
×25.7 κϑ
µολ ΝΗ 4ΝΟ3×
103 ϑ
κϑ= 1.61×10
4 ϑ is
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the heat absorbed by the dissolving ammonium nitrate. This is equal to
the heat absorbed by the water.
−1.61 ×10 4 J = 175 g H 2O ×4.184 J
g ⋅°C× (X °C − 25 °C)
X = 3.
1 °Χ
(c) It is assumed that there is no heat loss or gain between the heat pack and the
surroundings.
Section 7.7. Bond Enthalpies
7.28. Ethanol, by itself or in blends, is an important alternative fuel for gasoline-
powered engines. The molecular equation for the complete combustion of ethanol
is:
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
(a) Rewrite the balanced equation using full Lewis structures for each reactant
and product molecule.
(b) Use your model set and build a model of each reactant and product molecule
in this reaction.
(c) Refer to the Lewis structures, the models, and Table 7.3 to determine the
total enthalpy input that will be required to break all the bonds in the reactants.
(d) Refer to the Lewis structures, the models, and Table 7.3 to determine the
total enthalpy that will be released in forming all bonds in the products.
(e) What is the net enthalpy change in this combustion reaction? Is your result
likely to be an overestimate or underestimateof the experimental ∆ H oreaction?
Explain your reasoning.
(f) Ethanol is also a fuel when consumed by humans. Will your result in part (e)
also give an estimate for ethanol’s fuel value in our bodies? Why or why not?
Answer to 7.28:
(a)
Answer 7.1.
C
R
H
C
H
H
O +H H O O3 O C O2 + 3 O H
H
Answer 7.2. (b) Use your model set and build a model of each reactant
and product molecule in this reaction.
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Answer 7.3. (Could insert photo of models here)
Answer 7.4. (c) ∑ BH reactants = (5 × BH C–H + 1 × BH C-C + 1 × BH C-O + 1
× BH O-H) + (3 × BH O=O)
= [(5 mol × 414 kJ·mol-1) + (1 mol × 347 kJ·mol-1) + (1 mol × 351
kJ·mol-1
) + (1 mol × 460 kJ·mol-1
)] + [(3 mol × 499 kJ·mol-1
)= 4725 kJThis is the energy required to break all bonds in reactants. The sign is
positive showing bond breaking is an endothermic process.
(d) ∑ BH products = (2 × 2 × BH C=O + 3 × 2 × BH O-H)
= [(4 mol × –745 kJ·mol-1) + (6 mol × –460 kJ·mol-1) = –5740 kJ This is
the energy released in forming all bonds in products. The sign is negative
showing that bond formation is an exothermic process.
(e) ∆ H = ∑ BH reactants + ∑ BH products = +4725 kJ + –5740 kJ = –1015 kJ The
combustion of alcohol is an exothermic process overall, just as is expectedif the substance is being used as a fuel. The experimental reaction enthalpy
will involve a reactant (ethanol) and product (water) in their liquid states,
but the calculations based on bond enthalpies assume all species are gases.
Since this is an exothermic reaction, the ethanol is at a higher energy than
the products. Converting gaseous ethanol to liquid will lower its energy,
so the effect would be to reduce the experimental relative to calculated
energy. Converting gaseous water to liquid will lower its energy and theeffect would be to widen the energy gap between the reactants and
products and thus increase the experimental relative to calculated energy.
Which effect would predominate? Three moles of water are liquefied, but
only one mole of ethanol. The molar enthalpy of vaporization (or
condensation) of ethanol and water are about the same (Table 1.2 in
Chapter 1), so the water effect would be larger and the calculated energy
would likely be an underestimate. (The value calculated from standardenthalpies of formation – Section 7.8 – is about -1368 kJ·mol–1, which, as
predicted, is somewhat larger than the value, -1231 kJ·mol–1, from bond
enthalpies.)
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(f) The calculation of the enthalpy of combustion was done using average
bond enthalpies for gaseous compounds at 25°C and there is some error in
using such values for compounds in other states. There surely would also
be some differences if combustion through metabolic processes were to
take place. Still, values for food energy are determined through
combustion reactions carried out in typical calorimetric experiments.
7.29. Ethanol, CH3CH2OH, and dimethyl ether, CH3OCH3, are isomers. Use average
bond enthalpies to determine which of the isomers is the more stable. By how
much? Clearly explain your reasoning and the calculations you do to get your
result.
Answer to 7.29:
The more stable isomer will require greater energy to dissociate it into
atoms. We can use bond enthalpies to calculate the enthalpy change for
atomization of each isomer and compare the results.
ethanol dimethyl ether
bond # bonds BH, kJ # bonds BH, kJ
C-H 5 2070 6 2484
C-C 1 347
O-H 1 460
C-O 1 351 2 702
Total 3228 Total 3186
7.30. Ethanol requires more enthalpy to atomize to the same collection of atoms, so it is
the more stable of these two isomers by about 42 kJ (per mole of
isomer).Consider this reaction in which two amino acids join to form a peptide
bond, releasing a water molecule:
C C
H
N
R
H
H
O
O H
C C
H
N
R'
H
H
O
O H
C C
H
N
R
H
H
O
N C
H
C
O
O HR'
H
O
H
H
Use average bond enthalpies to estimate the enthalpy change for this reaction.
Answer to 7.30:
∆ H = ∑ BH reactants + ∑ BH products= (1 × BH C–O + 1 × BH N–H) + (1 × BH C–N + 1
× BH O–H)= [(1 mol × 347 kJ·mol-1) + (1 mol × 393 kJ·mol-1)] +
[(1 mol × 276 kJ·mol-1) + (1 mol × 460 kJ·mol-1)] = 740 kJ + 736
kJ = +4 kJ
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7.31. The glucose molecule can exist in a number of different forms, including the
open-chain and cyclic forms represented below. By convention, the carbons in thechain are numbered 1 through 6 beginning at the top, as shown. Carbon-1 in the
cyclic form is the one at the far right of the structure. To make the cyclic form, the
oxygen bonded to carbon-5 in the linear form becomes the ring oxygen in the
cyclic form where it is bonded to both carbon-1 and carbon-5. Which is the more
stable, the open-chain or the cyclic structure? How much more stable? Show your
reasoning and calculations clearly.
C
C
C
C
C
C
H
H
OHH
HHO
OHH
H
OHH
OH
1
2
3
45
6
O
C
CC
C
OC H
H
H
O
H
O
H
O
COH
HH
H
HO
H
H
open chain cyclicAnswer to 7.31:All the bonds in the two forms are the same, except those involved in
making the cyclic form from the open-chain form. The change involves
breaking a C=O on carbon-1 and making two C-O bonds in its place.
An O-H bond also has to be broken, but another is made, so they cancel out. Thus,
the difference is 745 kJ for C=O compared to 702 kJ for 2 C-O. By this
calculation, the open-chain form would be more stable (require more energy to
atomize) by about 43 kJ than the cyclic form. In nature, almost all glucose
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molecules exist in cyclic forms. There must be another stabilizing influence (or
bond enthalpy calculations for glucose, which is a solid, are not accurate enough
to give us the actual enthalpy difference).
7.32.Two isomers with the molecular formula CH2N2 are:
C N
H
H
NC
N
N
H
H
diazomethane diazirine
Diazomethane and diazirine are both gases. Under appropriate conditions, each
can react to give ethene and nitrogen:
2CH2N2(g) → H2C=CH2(g) + 2N2(g)
(a) Use average bond enthalpies to calculate ∆ H °reaction for diazomethane and for
diazirine undergoing this reaction. Are the reactions exothermic or endothermic?(b) Construct enthalpy-level diagrams (like Figure 7.11) for the two reactions.
Which isomer is the more stable? Are there any factors that might complicate use
of bond enthalpies for these molecules?
(c) Does our molecular bonding model help explain the relative stability?
Explain why or why not.
Answer to 7.32:
(a) Note that, for both reactions, we can ignore the C-H bonds, since there are
four in reactants and four in the products. We have to focus on the carbon-nitrogen, and nitrogen-nitrogen bonds.
For the reaction of two moles of diazomethane, we have:
∆ H rxn = – BH (C=C) –2 BH (N≡N) + 2 BH (C=N) + 2 BH (N=N)
∆ H rxn = – 620 – 2·(941.4) + 2·(615) + 2·(418) = –437 kJ
For the reaction of two moles of diazirine, we have:
∆ H rxn = – BH (C=C) –2 BH (N≡N) + 4 BH (C–N) + 2 BH (N=N)
∆ H rxn = – 620 – 2·(941.4) + 4·(276) + 2·(418) = –563 kJ
These reactions are highly exothermic with the predominant contribution in bothcases coming from the great stability of the triple-bonded nitrogen product. Note
that that only difference between these two calculations is the breaking of
different kinds of carbon-nitrogen bonds. For diazomethane, two C=N bonds are
broken and, for diazirine, four C–N bonds are broken. Recall that the enthalpy
change required to break one C=N bond is greater than twice the enthalpy change
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to break two C–N bonds. The difference, 63 kJ. For two moles of diazomethane
and diazirine, the difference in reaction enthalpies is 126 kJ (= 2·63 kJ), reflecting
the difference in carbon-nitrogen bonds broken.
(b)
Diazomethane is the more stable molecule, relative to diazirine. Complicatingfactors could be the strain of making the three-membered ring (which distorts the
usual bonding angles for carbon and nitrogen) in diazirine and/or the formal
separation of charge in the diazomethane molecule that could make it less stable
than a comparable molecule without formal charge.
(c) You might argue that the strain energy in diazirine is responsible for its higher
energy (63 kJ·mol–1 higher, based on these bond enthalpy calculations), but the
argument is not wholly satisfactory. To complicate matters, we find
experimentally that diazirine can be kept as a gas in the dark at room temperatureindefinitely, whereas samples of gaseous diazomethane react as shown here and
on the walls of their containers within a few hours. Based on exothermicity, the
reactions shown here seem very favorable, although we know that exothermicity
is not a good criterion for deciding the direction of a reaction. Diazomethane does
react, but some factor slows the diazirine reaction so much that it is undetectable.
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7.33. The table at the right gives the molecular
formulas and enthalpies of combustion for
three common sugars found in living
organisms. What conclusions can you drawfrom these data about the bonding in these
molecules? Explain your response.
Answer to 7.33:
As you see, these molecules are isomers, and their complete combustion will give
identical products in each case: 6 molecules of CO2 and 6 molecules of H2O for
each molecule of sugar oxidized. The identical values of their enthalpies of
combustion (to within 0.3%) indicates that the bonding must be essentially
identical in these isomers. That is, there must be the same number of C–C, C–H,
C–O, C=O, and O–H bonds in all three molecules. This is true if you compare
their open-chain forms with one another or their cyclic forms with one another (in
nature, the cyclic forms vastly predominate).
7.34. Disaccharides are sugars formed by the combination of two simpler sugars. A
maltose molecule, for example, is a combination of two glucose molecules:
C
C
C
C
OC H
H
H
O
H
O
H
O
COH
H H H
HO
H
H
C
C
C
C
OC H
H
H
O
H
O
H
O
COH
H H H
HO
H
H
C
CC
C
OC H
H
H
O
H
O
H
O
C
HH
H
HO
H
H
C
CC
C
O
C H
H
H
O
H
O
C
OH
HH
HO
H
H
O
H
H2
O
(a) Estimate ∆ H °reaction, using bond enthalpies. Is it necessary to do any numerical
calculations to make this estimate? Explain why or why not.
(b) Use your result from part (a) and the data from the Problem 7.33 to estimate
the enthalpy of combustion of maltose. The experimental value is –5644 kJ·mol –1.
How well does your estimate compare to the experimental value? How do you
explain the agreement or lack of agreement?
Answer to 7.34:
(a) In this reaction a C–O bond is broken in one sugar and an O–H bond in
the other. Then to make the products, a C–O bond is made between the
two sugars and an O–H bond is made to produce the water (from the OH
that had left a sugar). Exactly the same number and kind of bonds are
broken and made. Bond enthalpy calculations (which we do not have to
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Sugar Formula∆ H
combustion
kJ ⋅mol–1
fructose C6H12O6 –2812
galactose C6H12O6 –2803
glucose C6H12O6 –2803
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do, if there is no change in the kind and number of bonds) give an
estimated ∆ H reaction = 0 kJ for this reaction.
(b) One way to estimate the enthalpy of combustion of maltose is to break
the reaction into two reactions: hydrolysis of maltose to give two glucose
and then combustion of the glucose:
maltose + H2O → 2 glucose ∆ H rxn = 0 kJ
2 glucose + 12 O2 → 12 CO2 + 12 H2O ∆ H rxn = – 5606 kJ (from 7.17)
The sum of these two reactions is:
maltose + 12 O2 → 12 CO2 + 11 H2O ∆ H rxn = – 5606 kJ
This is the combustion reaction for maltose; we predict ∆ H rxn = – 5606 kJ to be
compared with the experimental value of –5644 kJ. The difference is small (only
about 0.6%), but the combustion enthalpies are experimental values that are good
to at least four significant figures. This means that the estimated value of 0 kJ for
the formation of maltose (or the reverse, its hydrolysis) is incorrect. The actual
value for the hydrolysis above must be about –36 kJ, in order to make the sum of
the two reaction enthalpies above equal to the measured enthalpy of combustion
of maltose. We know that bond enthalpies are averages that can be off by a few
percent for any particular bond in a molecule, so we shouldn’t be surprised that
this small discrepancy occurs. It is a warning, however, not to trust bond
enthalpies when the quantity of interest is the difference between two large values
-- small errors in the large values can lead to large percentage discrepancies in
their difference.
7.35. The reaction between hydrazine and hydrogen peroxide in rocket engines can be
represented as:
H2NNH2(g) + 2HOOH(g) → N2(g) + 4H2O(g)
(a) Use average bond enthalpies to estimate the standard enthalpy change (in kJ
per mole of hydrazine) for the reaction.(b) Compare the energy released in the hydrazine reaction to the energy that
would be obtained if two moles of ammonia were oxidized by hydrogen peroxide.
2NH3(g) + 3HOOH(g) → N2(g) + 6H2O(g)
(c) Which provides the most energy per gram of fuel, hydrazine or ammonia?
Answer to 7.35:
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(a) Let’s cancel the H–O bonds before going through all the arithmetic.
Four are broken on the left and eight are made on the right, so we will
include four in the products, but ignore them in the reactants. The ∆ H rxn
for this reaction is given by:
∆ H rxn = – BH (N≡N) –4 BH (H–O) + 4 BH (H–N) + BH (N–N) + 2 BH (O–O)
∆ H rxn = – 941 – 1840 +1572 + 193 + 284 = – 732 kJ
(b) In this case, canceling the H–O bonds leaves a net of six in the
products:
∆ H rxn = – BH (N≡N) –6 BH (H–O) + 6 BH (H–N) + 3 BH (O–O)
∆ H rxn = – 941 – 2760 + 2358 + 426 = –917 kJ
Two moles of ammonia reacting release about 25% more enthalpy than a
mole of hydrazine.
(c) A mole of hydrazine is 32 g of hydrazine, so the enthalpy change from
hydrazine is – 22.88 kJ·g–1 ((= –732 kJ)/(32 g)). The comparable value for
ammonia is –26.97 kJ·g–1 ((= –917 kJ)/(34 g)).
Since more enthalpy is released by the ammonia reaction, why isn’t ammonia
used, instead of hydrazine as the fuel? There are quite practical reasons for using
hydrazine. Probably the most important, from an engineering point of view, is that
hydrazine is a liquid at room temperature, Table 7.4, and can be easily stored and
pumped around in the rocket. Ammonia, on the other hand, is a gas at roomtemperature and has to be refrigerated and/or kept under high pressure to store it
as a liquid. This just increases the complexity of the rocket. Consider also the
oxidant, hydrogen peroxide. About one-and-one-half times as much peroxide is
required for the oxidation of ammonia as for an equal mass of hydrazine.
Therefore, the amount of fuel and oxidant together is less for hydrazine. The same
size rocket can carry a somewhat greater payload with the hydrazine fuel because
less mass of fuel and oxidant are needed. Work out for yourself the enthalpy
change for reaction of one gram of the stoichiometric mixture of fuel and oxidantfor each fuel.
7.36. WEB Chap 7, Sect 7.7.2-3. On Web Companion page 7.7.2 you select the
bonds that are broken in the reactants and made in the products.
(a) Use the data on page 7.7.3 to calculate the enthalpy change for this process.
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(b) How does your result in part (a) compare with the analysis on page 7.7.3
where all reactant bonds are broken and all product bonds formed?
(c) Explain why the comparison in part (b) comes out the way it does.
Answer to 7.36:
(a) The two bonds that have to be broken in the reactants and the three that haveto be made in the product are indicated here:
C
H
H
H C H
O
H
H
C
H
H
H C H
O
H
H
Breaking the H–O and C=C bonds requires 460 + 620 = 1080 kJ·mol–1. Making
the C–H, C–O, and C–C bonds releases energy, so the enthalpy change is
negative, – (414 + 351 + 347) = –1112 kJ·mol–1. The net change in enthalpy is the
sum of these two values, 1080 + (–1112) = –32 kJ·mol–1.
(b) On Web Companion page 7.7.3, you find that breaking all the bonds in the
reacatants requires 3196 kJ·mol–1 and making all bonds in the product releases –
3228 kJ·mol–1. The net change for the reaction is –32 kJ·mol–1. The results for the
two methods are identical.
(c) Finding that the results of the two methods are identical is expected. When all
the bonds in the reactants are broken (atomizing the molecules), we are breaking
several kinds of bonds, several C–H bonds, for example, that will simply be
remade in the product. Our assumption in using average bond enthalpies is that
the bond enthalpy for a given kind of bond is the same in all molecules. Thus
when the bonds are remade in the product, exactly the same amount of enthalpy isreleased as was required to break the bonds in the first place -- the breaking and
making simply cancel out. The only bond breaking and making that contribute to
a net difference for the reaction are those that are different in the reactants and
product, which is what we focused on in part (a). You can see this argument
graphically on Web Companion page 7.7.3, if you first break the two bonds
selected on page 7.7.2 and then break all the others to atomize the reactants. Now
when you make the product bonds, first make all those that are the same in the
product as in the reactants and then, finally, make the three new bonds. Comparethe lengths of the bars on the graph for the breaking and making the same set of
bonds.
7.37. Carbon–hydrogen bond-dissociation enthalpies are shown for two different
hydrogens in the reactions of propene and 2-butene shown here. The enthalpies
for the two different bond dissociations are quite different. The dissociation of a
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carbon–hydrogen bond on a carbon that is next to a doubly bonded carbon,
reactions (1) and (3), requires less energy.
CH
H
H
C C
H
H
H
CH
H
H
C C
H
C
H
H
H
H
CHH
C CH
H
H
CH
H
H
C CH
H
CH
H
C C
H
C
H
H
H
H
CHH
H
C C
H
C
H
HH H
H
H
H
dissociation enthalpy
kJ mol–1
361
430
358
430
(1)
(2)
(3)
(4)
(a) Are the propyl and butyl free radicals formed in reactions (1) and (3),
respectively, more stable (lower enthalpy) or less stable (higher enthalpy) than
those formed in reactions (2) and (4)? Clearly explain the reasoning for your
response, using enthalpy level diagrams, if these aid your explanation.
(b) The unpaired electron in the three-carbon and four-carbon free radicals
formed in reactions (1) and (3) is often shown as participating in delocalized π
bonding (see Chapter 5, Section 5.7), with the two π electrons from the adjacent
double bond. Write Lewis structures supporting this model. Do the bond-
dissociation enthalpies and your interpretation in part (a) support this model?
How would delocalization of the unpaired electron orbital affect the energy of the
radical? Explain your reasoning clearly.
Answer to 7.37:
(a) The enthalpy level diagram at the right represents the enthalpies of the
reactant and the two sets of products from reactions (1) and (2).
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It takes less enthalpy to produce the
products from reaction (1). Thus, the free radicals formed in reaction (1)
{and (3)} are more stable (by 69 kJ) than those formed in reaction (2)
{and (4)}.
(b) The “equivalent” Lewis structures for the three- and four-carbon free radicals
formed in reactions (1) and (3) are:
CH
H
C C
H
H
HC H
H
CC
H
H
HCH
H
C C
H
C
H
H
H
H CH
H
C C
H
C
H
H
H
H
The structures for the three-carbon radical are identical in energy, so we know
from our discussion in Chapter 5, Section 5.7, that the electrons involved are
delocalized and of lower energy than they would be if the two pi double-bond
electrons and single free-radical electron did not interact. The two four-carbon
structures are not quite identical, but the sigma bonding to the carbon atoms with
the free radical electron is very similar in the two structures (to two hydrogen
atoms in the first and to a hydrogen atom and carbon atom in the second), so we
can treat the structures as energy-equivalent and again see that there is electron
delocalization to lower the energy of this free radical
Section 7.8. Standard Enthalpies of Formation
7.38. You usually see sulfur in the form of a light yellow powder, the rhombic
crystalline allotrope of sulfur. If you melt this solid (m. p. 112.8 °C), pour the
liquid into a filter paper cone, and then unfold the cone as the sample cools, you
find that the liquid has solidified in dark yellow needles. This is the triclinic
crystalline allotrope of sulfur. After several days, the needles become covered by
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a light yellow layer. Which allotrope of sulfur is probably its most stable form
under standard conditions? Explain your answer.
Answer to 7.38:
The rhombic allotrope, the light yellow powder, is the more stable at room
temperature. At higher temperatures, near the melting point, the triclinic allotrope
must be more stable, since it is the form that crystallizes from the molten sulfur.
Upon standing at room temperature, the rhombic allotrope forms slowly on the
surface of the triclinic needles, which indicates a slow transformation to the more
stable form. If we wait long enough, all the needles change back to the light
yellow powder.
7.39. Urea (Latin urina = urine), H2NC(O)NH2, is a water-soluble compound made by
many organisms, including humans, to eliminate nitrogen. Use standard
enthalpies of formation to find the enthalpy for the reaction producing urea from
ammonia and carbon dioxide.
2NH3(g) + CO2(g) → H2NC(O)NH2(s) + H2O(l)
Show all your work so your procedure is clear.
Answer to 7.39:
The standard enthalpies of formation for ammonia, carbon dioxide, urea,
and water are – 46.1, –393.5, –333.0, and –285.8 kJ·mol–1, respectively.
The standard enthalpy change for the reaction is (where the stoichiometric
coefficients -- units of mol -- are not shown, if they are unity):
∆ H orxn = ∆ H of (H2O(l)) + ∆ H
of (H2NC(O)NH2(s)) – 2∆ H
of (NH3(g)) – ∆ H
of (CO2(g))
∆ H orxn = (–285.8 kJ) + (–333.0 kJ) – (– 92.2 kJ) – (– 393.5 kJ) = –133.1 kJ
7.40. In Problem 7.39, you used enthalpies of formation to calculate the enthalpy for
making urea from ammonia and carbon dioxide. How could you estimate the
enthalpy of vaporization of urea? Explain the procedure you would use and then
carry it out. Hint : What other way could you estimate the enthalpy of the reaction
forming urea? How do the two ways differ?Answer to 7.40:
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Recall that bond enthalpy values are
applicable to gas phase molecules. We
could use bond enthalpies to calculate the
enthalpy change for the reaction in 7.25
with the urea and water as gases and
compare the result with the enthalpy
change calculated in 7.25 for the urea as a
solid and water as a liquid. The difference
is an estimate of the enthalpy of
sublimation of urea plus the enthalpy of
vaporization of water. (The problem is
misstated when it asks for the enthalpy of
vaporization of urea. You can get the
enthalpy of sublimation, but have no way
to estimate the changes from solid to
liquid to gas.) For the gas phase reaction
we have:
∆ H orxn = –2 BH (H–O) –4 BH (H–N) – 2 BH (C–N) – BH (C=O) +
6 BH (H–N) +
2 BH (C=O in CO2)
∆ H orxn = – 920 – 1572 – 552 – 745 + 2358 + 1598 = + 167 kJ
Enthalpy is required to make gaseous urea and water from ammonia and carbon
dioxide. The enthalpy change (decrease -- exothermic change) going from the
gases to solid urea and liquid water is -133 – (167) = – 300 kJ. The enthalpy of
vaporization of water is 44 kJ·mol–1, so the condensation of water accounts for –
44 kJ. This leaves – 256 kJ as the enthalpy released when gaseous urea condenses
to solid urea. Our estimate for the enthalpy of sublimation of urea is about 256
kJ·mol–1.
7.41. During fermentation of fruit and grains, glucose is converted to ethanol and
carbon dioxide according to this reaction:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
(a) Using the data in Appendix XX, calculate ∆ H °reaction.
(b) Is this reaction exothermic or endothermic?
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(c) Which has higher enthalpy, the reactant or the products?
(d) Calculate ∆ H for the formation of 5.0 g of C2H5OH(l).
(e) What quantity of heat is released when 95.0 g of C2H5OH(l) is formed at
constant pressure?
Answer to 7.41: (a) ∆ H °reaction = -68.0 kJ•mol-1 (b) exothermic (c) reactants
(d) -3.7 kJ (e) -70.2 kJ
7.42. Consider this gas phase reaction in which methanoic (formic) acid reacts with
ammonia. The product formamide contains a peptide bond and there is the release
of a water molecule:
H C
O
O H
HN
H
H
H C
O
N H
H
O
H
H
(a) Use average bond enthalpies to estimate the enthalpy change for this
reaction. How does this ∆ H ° compare to the ∆ H ° you calculated in problem 7.29?
Would you have expected this result? Why or why not?
(b) Use standard enthalpies of formation to calculate the change in enthalpy for
this reaction, assuming reactants and products are gases in their standard states.
The enthalpies of formation of gaseous methanoic acid and formamide are –379
and –186 kJ·mol–1, respectively.
(c) Compare the results of your calculations in parts (a) and (b). Offer some
reasons to help explain your comparison.
Answer to 7.42:
(a) ∆ H = ∑ BH reactants + ∑ BH products= (1 × BH C–O + 1 × BH N–H) + (1 × BH C–N +
1 × BH O–H)= [(1 mol × 347 kJ·mol-1) + (1 mol × 393 kJ·mol-1)] + [(1 mol
× 276 kJ·mol-1) + (1 mol × 460 kJ·mol-1)] = 740 kJ + 736 kJ= +4 kJ
Note that the calculation using average bond enthalpies is the same as in
the previous problem, for exactly the same bonds are broken and formed.
(b) Calculation using standard enthalpies of formation.
∆ H reaction = Σn p∆ H 0
f (products) – Σnr ∆ H 0
f (reactants)
= [(1 mol) × –254 kJ·mol-1) + (1 mol) × –285.8 kJ·mol-1)] –
[(1 mol) × –363 kJ·mol-1) + (1 mol) × –46.3 kJ·mol-1)]= –130.5 kJ
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(c) The two calculations are not very close. Average bond enthalpies are based on
gaseous compounds at 25 °C. Standard enthalpies of formation are based on the
assumption that reactants and products are in their standard states at 25 °C.
Neither set of assumptions can be expected to hold true for this reaction, which
likely takes place in aqueous solution. The bond enthalpies within the water
molecule are given for the gaseous state, but the standard state for water at 25 °C
is liquid. Even correcting for the Enthalpy of Vaporization for water does not
yield that close a result, because it still would be necessary to correct for the
Enthalpy of Solution values for the other reactants and products.
7.43. The combustion of ammonia is represented by this equation:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Experimentally, we find ∆ H °reaction = – 905 kJ for the reaction as written.
(a) Use the standard enthalpies of formation for NO(g) and H2O(g) from
Appendix XX and the standard enthalpy change for the reaction to calculate the
standard enthalpy of formation for NH3(g).
(b) How does your result in part (a) compare with the standard enthalpy of
formation for NH3(g) in Appendix XX? Explain why they are the same (or
different).
Answer to 7.43:
(a) ∆Hreaction = Σnp∆H0f (products) – Σnr ∆H
0f (reactants)
–905 kJ = [(4 mol × 90.4 kJ·mol-1) + 6((1 mol × –241.8 kJ·mol-1)] –
[4(∆ H 0f (NH 3) + 5(0)]
–905 kJ = [361.6 kJ – 1450.8 kJ] – 4(∆ H 0f ( NH 3)
∆ H 0f ( NH 3) = –46.0 kJ·mol-1
(b) The table value is –46.3 kJ·mol-1. This is excellent agreement and points up
the fact that the table values are often determined by means of values for
combustion reactions which are relatively straightforward to run.
7.44. In the process known as coal gasification, coal can be reacted with steam and
oxygen to produce a mixture of hydrogen, carbon monoxide, and methane gases.
These gases are desirable as fuels and they can also serve as the starting material
for the synthesis of other organic substances such as methanol, used for the
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production of synthetic fibers and plastics. Consider these three reactions that can
take place in the process of coal gasification.
Reaction 1: C(s) + H2O(g) → H2(g) + CO(g)
Reaction 2: C(s) + 1 / 2O2(g) → CO(g)
Reaction 3: C(s) + 2H2O(g) → CH4(g) + O2(g)
Use standard enthalpies of formation to find the enthalpies of reaction for each of
these three reactions. If it is possible to control the reaction conditions to favor
one or more of these reactions, which is the most energetically favorable? the
least energetically favorable? Explain your reasoning.
Answer to 7.44:
(a) Reaction 1: ∆ H orxn = +373.1 kJ
Reaction 2: ∆ H orxn = –110.5 kJ
Reaction 3: ∆ H orxn = –393.5 kJ
(b) The most exothermic reaction is Reaction 3. This does not imply that chemists
should work to control the reaction conditions to maximize this reaction. The
main goal of coal gasification is to produce fuels and starting materials for
synthesis. CO and H2 are the desired products, but the reaction to produce carbon
dioxide takes place at the same time.
Section 7.9. Harnessing Energy in Living Systems
7.45. What is a coupled reaction? What is its importance in biological reactions?
Answer to 7.45:
In a coupled reaction, the reaction providing the energy does not proceed unless
the reaction requiring the energy occurs. For example, the exothermic oxidation
of glucose is coupled with the endothermic formation of adenosine diphosphate,
ADP3-, from adenosine triphosphate, ATP4-. The reverse reaction, the hydrolysis
of ATP4- to ATP3- is exothermic and is coupled to other energy-requiring
biological functions such as locomotion, information processing, and synthesizingnew biological molecules.
7.46. What is the role of ATP in biological reactions?
Answer to 7.46:
The function of ATP is to store energy when energy is not needed for biological
reactions and to release it instantly when biological reactions require it.
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7.47. Coupled reactions take place in living systems, but coupled processes have many
applications in industry as well. For example, waste heat from a power plant can
be captured and used to do work, such as desalting seawater.
(a) Other than the availability of seawater, what do you think might limit the
usefulness of this example of a coupled process?
(b) Can you think of any other examples of energetically coupled processes from
your personal experience or other courses you have taken? Briefly describe such a
process. You might find it useful to research this topic using web-based resources.
Answer to 7.47:
(a) There are limitations on the overall efficiency of this coupled process.
One of the major limiting factors is using waste heat from a power plant
for desalting seawater is that the heat content of the waste heat is relatively
low. It is not great enough to boil water, so a process of flash distillation is
used. This coupled process is not widely used in the world, although solar
distillation processes are more common.
(b) Answers will vary depending on the background of the students. They may
present examples from engineering, other biological applications, or from ecology
courses. Note: This question could be assigned as a web-based research question.
7.48. Many ATP4--coupled reactions require that Mg2+ be present as well. What do you
think is the function of Mg2+ in these reactions? Hint : See Chapter 6, Section 6.6.
Answer to 7.48: Mg2+ can complex with the negatively charged ATP ions,
providing a three dimensional scaffold that allow these coupled reactions to occur.
7.49. The aerobic reaction sequence of glycolysis involves the complete oxidation of
pyruvic acid, CH3C(O)C(O)OH (= C3H4O3). This reaction is coupled to the
formation of ATP4-, according to this overall reaction:
2C3H4O3(l) + 5O2(g) + 30ADP3-
(aq) + 30HOPO32-
(aq) + 30H+(aq) →
6CO2(g) + 34H2O(l) + 30ATP4-
(aq)
(a) Calculate the ∆H° for the uncoupled oxidation of pyruvic acid toCO2 and H2O.
(b) Calculate the ∆H° for the coupled reaction and determine the
percentage of energy converted to ATP!.
Answer to 7.49:
(a) The separate equation for the uncoupled oxidation of pyruvic acid is:
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2C3H4O3(l) + 5O2(g) 6CO2(g) + 4H2O(l) ∆H°pyruv oxid "
!2##2 k$
∆H°pyruv oxid i% calculated from the heat% for formation a% %hown&
∆H°pyruv oxid " '(mol)!#*#.+ k$•mol!- )mol-)!2/+./ k$•mol!-0
! '2mol)!+/+./ k$•mol!-
" !2##2 k$
(b) 1ecall that ∆H°reaction" 2 k$•mol! for the formation of
one ATP4- from ADP3- and HOPO32-. Therefore, the energy required
to form 30 ATP4- will be 30 times as great.
30ADP3-(aq) + 30HOPO32-
(aq) + 30H3O+(aq) 60H2O(l) + 30ATP
4-(aq)
∆H°reaction " )#mol-)2 k$•mol!- " (# k$
The ∆H°coupled reaction i%& )!2##2 k$- (#k$ " !3#k$.
The 234 percent e5ciency i% calculated a% follow%&
Only 630kJ of the 2332kJ released in this oxidation are used effectively in the
production of ATP4-. The remaining 1703kJ is lost as heat.
Section 7.10. Pressure-Volume Work, Internal Energy, and Enthalpy
7.50. An inventor claims to have built a device that is able to do about 0.8 kJ of work
for every 1 kJ of thermal energy put into it. He is looking for investors to buy
shares in his company, so he can commercialize his invention. Would you invest
in his company? Explain why or why not.
Answer to 7.50:
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You should, by all means, invest in this inventor (after you have seen proof that
his device can actually do what he says). This device is about 80% efficient in
converting random thermal motion to the directed motion of work. This is a good
deal better than the best engines now available. There is no violation of the first
law here, since he is not getting out more than he puts in. The “lost” energy
presumably remains in the device in some form or is expelled as thermal energy
to some other part of the surroundings. The inventors you have to look out for are
those who claim to get more work out of a thermal engine than the thermal energy
they put in; that is a violation of the first law.
7.51. The volume of a gas is decreased from 10. L to 1.0 L at a constant pressure of 5.0
atmospheres. Calculate the work associated with the process. Is work done on the
gas or by the gas? (1 L·atm = 101.3 J)
Answer to 7.51:
(a) P∆V = (5 atm) (10.0 L - 1.0 L) = 45 L·atm; 45 × 101.3 J
1 L ⋅ατµ= 4.6 × 103 J
(b) Work is being done on the gas, as the volume is decreasing from 10.0 L to 1.0
L. A positive sign is associated with work done on the system.
7.52. A hot air balloon is inflated by using a propane burner to heat the air in the
balloon. If during this process 1.5 × 108 J of heat energy cause the volume of the
balloon to change from 5.0 × 106 L to 5.5 × 106 L, what are the values of q, w, and
∆ E for this process? Assume that the balloon expands against a constant pressure
of 1.0 atmosphere. (1 L·atm = 101.3 J)
Answer to 7.52:
The value of q p is given, this is 1.5 × 108 J. This is the heat energy added
to the system at constant pressure.
The pressure-volume work done can be calculated in this manner.
w = P∆V
w = (1.0 atm) (5.5 × 106
L – 5.0 × 106
L)
w = (1.0 atm) (5.5 × 106
Λ 7 5.0 × 106 Λ) ×
101.3 ϑ
1 Λ8ατµ= 5× 10
7ϑ
Finally, ∆ E can be calculated from the values for q p and w; ∆ E = q p – P∆V
∆ E = (1.5 × 108 J) – (5 × 107 J) = 1.0 × 108 J
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7.53. Photographic flash bulbs, once more common than they are now, produced light
from the reaction of either zirconium or magnesium wire with oxygen in the bulb,
for example:
2Mg(s) + O2(g) → 2MgO(s) + energy (light and heat)
They became quite warm when used, but only rarely did they actually explode!
Discuss the changes in ∆ E , q, and w for the chemical reaction that takes place in a
flash bulb.
Answer to 7.53:
The chemical energy of reaction of the metal with oxygen is transformed to light
energy and heat energy in the course of the reaction. As long as the bulb does not
explode, this is a constant volume reaction. This means no pressure-volume work
is possible, so w = 0. This also means ∆ E = ∆ H .
7.54. A gaseous chemical reaction occurs in which the system loses heat and contracts
during the process. What are the signs, "+" or "-", for P∆V , ∆ E and ∆ H ? Explain
your reasoning.
Answer to 7.54:
Since ∆ H = ∆ E + P∆V , we know that ∆ H is negative since the system loses heat.
The system contracts so Vfinal < Vinitial. This means that P∆V will be negative since
P is always a positive number. ∆ E may be either positive or negative depending
on the magnitude of ∆ H and P∆V .
7.55. Consider two experiments in which a gas is compressed in a closed syringe by
pushing the plunger against the trapped gas. Both experiments use the same size
syringe.
Experiment A Experiment B
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(a) How does the pressure-volume work performed in Experiment A compare to
that performed in Experiment B? Assume the external pressure is the same in both
experiments. Explain your reasoning.
(b) Assume that a third experiment is performed using a syringe with a wider
diameter than the one used in the first two experiments. How would the pressure-
volume work compare with that in Experiments A if ∆ x is the same? Assume the
pressure is the same in both experiments.
Answer to 7.55:
(a) The pressure-volume work performed in Experiment A is less than that
performed in Experiment B. Given that the cross-sectional area is the same
for both syringes and the pressure is constant, the pressure-volume work is
directly proportional to the value of∆ x
, which is smaller for ExperimentA. The mathematical relationship is:
w = P × A × ∆ x
(b) Although ∆ x is the same for both experiments, ∆V is not the same because the
cross-sectional area of the syringe is greater in the third experiment. Because
pressure -volume work is a product of the pressure, which is constant, and the
change in volume, the work done in the third experiment is greater than that done
in Experiment A.
7.56. Consider the changes that occur in this reaction:
#H2(g) 92(g) → 29H#(g)
Is work done by the system (gaseous reactants and product) or on the system?
Explain your answer. Assume that this reaction occurs at constant temperature
and pressure.
Answer to 7.56:
We calculate ∆V by noting that 3 liters of H2(g) react with 1 liter of N2(g) to
produce 2 liters of NH3(g). In this case, ∆V = -2 liters and work is greater than
zero, indicating that work must enter the system.
7.57. Most of the nitrogen used to inflate an automobile airbag is produced by the
reaction of sodium azide:
2[Na+][N3–](s) —> 2Na(g) + 3N2(g)
A driver’s side airbag might typically contain about 95 g of sodium azide.
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(a) A mole of nitrogen gas occupies about 25 L at 25 °C. How large is the airbag
this nitrogen will inflate?
(b) How much work is done by the nitrogen as it inflates the airbag?
(c) If this exothermic reaction is investigated in a constant volume calorimeter,
will the measured thermal energy release be greater or less than the thermal
energy released at constant pressure? How large will the difference be? Clearly
explain your reasoning.
Answer to 7.57:
(a) The stoichiometric equation shows that 3 moles of nitrogen gas are
produced for every two moles of the azide that reacts. The azide has a
molar mass of 65 g. Therefore:
mol N2 formed =95 g azide
65 g ⋅µολ17
?
??
?
??
3 µολ Ν2
2 µολ αζιδε
?
?
?
?= 2.2 mol N2
The volume occupied by 2.2 mol N2 is 55 L (2.2 mol × 25 L·mol–1) so the
airbag is about 55 L in volume, which is approximately 2 cubic feet (or
you can imagine about 25 empty, two-liter soft drink bottles suddenly
appearing in front of you when the airbag deploys).
(b) The external pressure against which the airbag pushes to inflate is
atmospheric pressure, Patm, which is 1.0 × 105 Pa (N·m–2). The volume
change, ∆V , of the gas is 55 × 10–3 m3 (since we go from no gas to a final
volume of 55 L and a liter is 10–3 m3 = a cubic decimeter). Thus:
work done by the gas = Patm·∆V = (1.0 × 105 N·m–2)·(55 × 10–3 m3) = 5500 J
(c) We know that ∆ H = ∆ E +P·∆V and that the pressure-volume product is a
positive 5500 J. Therefore, the measured thermal energy release at constant
volume, qV = ∆ E , will be smaller by 5500 J than that measured at constant
pressure, qP = ∆ H .
7.58. For this reaction, as written, ∆ H ° = -484 kJ:
2H2(g) + O2(g) → 2H2O(g)When 0.5 mol of H2(g) was reacted with 0.3 mol of O2(g), at a constant pressure of
1.0 atm, the change in volume was –6.1 liters. Calculate how much work was
done and the value of ∆ E ° for this reaction. Is work done on the system or by the
system?
Answer to 7.58:
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Since ∆ E = ∆ H - P∆V , we will first need to calculate both ∆ H and P∆V to
obtain the value for ∆ E . Note that the initial amounts of H2(g) and O2(g)
are not in the 2:1 molar ratio demonstrated by the balanced equation. In
fact, H2(g) is the limiting reagent and is consumed completely while
0.05mol of O2(g) remains unreacted. ∆ H for the amounts given is:
:ork )!Pext ∆V - i% determined next&
9ote& :ork i% done on the %y%tem.
Therefore, ∆E " !2k$ .(2k$ " 2.k$
Section 7.11. What Enthalpy Doesn’t Tell Us
7.59. Indicate whether each of the following statements is true or false. If a statement is
false, write the correct statement.
(a) Consider a 10 mL sample of pure water at 25 °C and 1 atm pressure (state
A). The sample is cooled to 1 °C and then the pressure is reduced to 0.5 atm
(state B). It takes 5 hours to carry out the change from state A to state B. The
sample is then heated and the pressure is raised to 1 atm. In one minute the water
is back at 25 °C ( the sample is back at state A). The internal energy change in
going from state A to B is equal to, but opposite in sign to the internal energy
change going from state B to A.
(b) The work done in changing from state A to B and the work done in changing
from state B to A in part (a) are numerically the same but opposite in sign.
(c) The enthalpy change for a change of state of a system is independent of the
exact state of the reactants or products.
(d) At constant pressure the amount of heat absorbed or evolved by a system is
called the enthalpy change, ∆ H .
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2mol H2)g- "
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(e) If volume does not change, the amount of heat released during a change of
state of a system is equal to the decrease in internal energy of that system.
(f) If a reaction is spontaneous, it is always exothermic.
(g) If the enthalpy change for this reaction, N2(g) + O2(g) → 2NO(g), is 180.5 kJ,
then the enthalpy change for this reaction, 1/2N2(g) + 1/2O2(g) → NO(g), is 90.2
kJ.
Answer to 7.59:
(a) True. Internal energy is a state function. The difference in the internal
energies of two states is independent of the path taken between the states.
(b) False. Work depends on the path, i.e., the method of going from one
state to another.
(c) False. The enthalpy of reaction is dependent on the exact state of the
reactants or products.
(d) True. The relationship between ∆ H and ∆ E is ∆ H = ∆ E +p∆V . At
constant volume the amount of heat absorbed or evolved by a system is
equal to eternal energy change.
(e) True. At constant volume ∆H = ∆E and qv= qp
(f) False. Endothermic reaction which are spontaneous include freezing
water at temperatures below 0oC and dissolving many salts (for example
NH4NO3).
(g) True.
7.60. When 1 mol of NH4Cl(s) is dissolved in water, forming NH4+(aq) and Cl-(aq), the
change in enthalpy is 14.8 kJ·mol-1. When 1 mol of CaCl2(s) is dissolved in water,
forming Ca2+(aq) and Cl-(aq), the change in enthalpy is –82.9 kJ·mol-1.
(a) :hich >eaker felt cool to touch after the %alt di%%olved in
water? :hich one felt warm?
(b) :hat would you have to do to maintain the temperature of
each >eaker at 2+°C?
(c) To determine ∆H° for di%%olution of %alt% in water, what
factor% mu%t you con%ider?
Answer to 7.60:
(a) The NH4Cl solution is cool to touch, while the CaCl2 solution is warm
to touch.
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(b) For the NH4Cl solution, 15.1 kJ of heat must be added. For the CaCl2
solution, 82.9 kJ must be removed.
(c) The concentration of the dissolved salt must be considered. The above values
for the enthalpy changes are obtained by measuring enthalpy changes and
extrapolating to the limit of infinite dilution. Also, we note that it is impossible to
assign the enthalpy changes associated with the individual ions of a salt without
giving an arbitrary enthalpy value to one of the ions.
7.61. Given that ∆ H °f for Cl-(aq) = –167.4 kJ·mol-1, use the enthalpy of reaction data
from Problem 7.60 to calculate ∆ H °f for NH4+(aq) and for Ca2+(aq).
Answer to 7.61:
For NH4+(aq):
∆H° form'9H)a@-0 ∆H° form'Cl!)a@-0 " ∆H° form'9HCl)%-0 +.
k$•mol!
∆H° form'9H)a@-0 )!(3. k$•mol!- " !#+. k$•mol! +.
k$•mol!
∆H° form'9H)a@-0 " !#2.* k$•mol!
or Ca
2
)a@-&
∆H° form'Ca2)a@-0 ∆H° form'2Cl
!)a@-0 " ∆H° form'CaCl2)%-0 ! /2.*
k$•mol!
∆H° form'Ca2)a@-0 ! ##./ k$•mol! " !3*+./ k$•mol! ! /2.*
k$•mol!
∆H° form'Ca2)a@-0 " !+#.* k$•mol!
7.62. A nineteenth-century chemist, Marcellin Berthelot, suggested that all chemical
processes that proceed spontaneously are exothermic. Is this correct? Give some
examples that justify your answer.
Answer to 7.62:
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Berthelot’s suggestion was incorrect. Endothermic processes that are spontaneous
include freezing water at temperatures below 0oC and dissolving many salts (for
example NH4NO3 ).
Section 7.13. EXTENSION — Ideal Gases and Thermodynamics
7.63. Consider two samples of gas in identical size containers. Which of these
statements is true? Explain your reasoning for each choice. Rewrite the false
statements to make them true.
(a) If the temperature of the two samples is the same, then the pressure of gas in
each container is the same.
(b) If the temperature and pressure of the two samples are the same, then the
number of moles of gas in each container is the same.
(c) If the gas in each container is the same and the temperature and number of
moles of gas in each container are the same, then the number of collisions with
the wall per unit time is the same in both containers.
Answer to 7.63:
(a) The statement, as given, is not true. Although the temperature and volume of
the gas samples are the same, we have no information about the quantity of gas,
that is, the number of moles of gas, in each container. The one with more moles
will have the higher pressure, since P = n( RT / V ) and all the quantities inside the
parentheses on the right are the same for both samples. We can amend the
statement in two ways: (1) If the temperature and number of moles of gas is thesame, then the pressure of gas in each container is the same or (2) If the
temperature of the two samples is the same, then the pressure will be higher in the
container with the greater number of moles of gas.
(b) This statement is true, since n = PV / RT and all the quantities on the right are
the same for both samples.
(c) This statement is true. To see why this is so, we note that the pressure of the
gas in each container is the same, since P = nRT / V and all the quantities on the
right are the same for both samples. Pressure is a result of collisions of the gas
molecules with the walls. Since the gas molecules an