.
5
Thermochemistry
Visualizing Concepts
5.1
(a)
Analyze/Plan. The exercise gives the charges and separation of two particles.
Use Equation 5.2 to calculate electrostatic potential energy, E e l. Solve.
el 
el 
 Q 1Q 2
d
;   8.99  109 J-m /C 2 ; Q 1 = Q 2 = 2.0  10–5 C; d = 1.0 cm
8.99  109 J-m
C2

100 cm
1

 2.0 105 C  2.0 105 C  359.6  3.6 102 J
1.0 cm
m
(b)
The spheres are both positively charged, so they will move away from each other.
(Like charged particles repel, oppositely charged particles attract.)
(c)
As the like charged spheres move apart, the electrostatic potential energy of the
system is converted to kinetic energy. As the distance between them approaches
infinity, potential energy approaches zero and the kinetic energy of each particle
is 1.8  102 J, one half of the initial potential energy calculated in part (a).
E k = 1/2 mv 2 ; v = (2 E k /m) 1/2; E k = ½(3.6  102 J) = 1.8  102 J
1/2

1.8  102 kg-m 2
1 
v   2


2

1.0 kg 
s

5.2
 18.97  19 m /s
(a)
The caterpillar uses energy produced by its metabolism of food to climb the twig
and increase its potential energy.
(b)
Heat, q, is the energy transferred from a hotter to a cooler object. Without
knowing the temperature of the caterpillar and its surroundings, we cannot
predict the sign of q. It is likely that q is approximately zero, because a small
creature like a caterpillar is unlikely to support a body temperature much
different from its environmental temperature.
(c)
Work, w, is the energy transferred when a force moves an object. When the
caterpillar climbs the twig, it does work as its body moves against the force of
gravity.
(d)
No. The amount of work is independent of time and therefore independent of
speed (assuming constant caterpillar speed).
(e)
No. Potential energy depends only on the caterpillar’s position, so the change in
potential energy depends only on the distance climbed, not on the speed of the
climb.
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5 Thermochemistry
5.3
5.4
5.5
5.6
Solutions to Exercises
(a)
The internal energy, E, of the products is greater than that of the reactants, so the
diagram represents an increase in the internal energy of the system.
(b)
E for this process is positive, (+).
(c)
If no work is associated with the process, it is endothermic.
(a)
For an endothermic process, the sign of q is positive; the system gains heat. This
is true only for system (iii).
(b)
In order for E to be less than 0, there is a net transfer of heat or work from the
system to the surroundings. The magnitude of the quantity leaving the system is
greater than the magnitude of the quantity entering the system. In system (i), the
magnitude of the heat leaving the system is less than the magnitude of the work
done on the system. In system (iii), the magnitude of the work done by the
system is less than the magnitude of the heat entering the system. None of the
systems has E < 0.
(c)
In order for E to be greater than 0, there is a net transfer of work or heat to the
system from the surroundings. In system (i), the magnitude of the work done on
the system is greater than the magnitude of the heat leaving the system. In
system (ii), work is done on the system with no change in heat. In system (iii), the
magnitude of the heat gained by the system is greater than the magnitude of the
work done on the surroundings. E > 0 for all three systems.
(a)
No. This distance traveled to the top of a mountain depends on the path taken by
the hiker. Distance is a path function, not a state function.
(b)
Yes. Change in elevation depends only on the location of the base camp and the
height of the mountain, not on the path to the top. Change in elevation is a state
function, not a path function.
(a)
State B
(b)
EAB = energy difference between State A and State B.
EAB = E 1 + E 2
(c)
or
EAB = E 3 + E 4
ECD = energy difference between State C and State D.
ECD = E 2 – E 4
or
ECD = E 3 – E 1
(Note that the sign of E depends on the definition of initial and final state, but
the magnitude is the absolute value of the difference in energy.)
5.7
(d)
The energy of State E is E 1 + E 4 , whereas the energy of State B is E 1 + E 2 .
Because E 4 > E 2 , State E is above State B on the diagram; State E would be the
highest energy on the diagram.
(a)
You, part of the surroundings, do work on the air, part of the system. Energy is
transferred to the system via work and the sign of w is (+).
(b)
The body of the pump (the system) is warmer than the surroundings. Heat is
transferred from the warmer system to the cooler surroundings, and the sign of q
is (–).
(c)
The sign of w is positive, and the sign of q is negative, so we cannot absolutely
determine the sign of E . It is likely that the heat lost is much smaller than the
work done on the system, so the sign of E is probably positive.
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5 Thermochemistry
5.8
5.9
5.10
5.11
5.12
Solutions to Exercises
(a)
The temperature of the system and surroundings will equalize, so the temperature of
the hotter system will decrease, and the temperature of the colder surroundings will
increase. The system loses heat by decreasing its temperature, so the sign of qsys is (–).
The surrounding gains heat by increasing its temperature, so the sign of qsurr is (+).
From the system’s perspective, the process is exothermic because it loses heat.
(b)
If neither volume nor pressure of the system changes, w = 0 and E = q = H.
The change in internal energy is equal to the change in enthalpy.
(a)
w = – PV. Because V for the process is (–), the sign of w is (+).
(b)
E = q + w. At constant pressure, H = q. If the reaction is endothermic, the signs
of H and q are (+). From (a), the sign of w is (+), so the sign of E is (+). The
internal energy of the system increases during the change. (This situation is
described by the diagram (ii) in Exercise 5.4.)
(a)
N 2 (g) + O 2 (g)  2 NO(g). Because V = 0, w = 0.
(b)
The reaction of two elements to form one mole of a compound fits the definition
of a formation reaction. Find the value for enthalpy of formation of NO(g) in
Appendix C. H = Hf = 90.37 kJ for production of 1 mol of NO(g).
(a)
H A = H B + H C . Diagram (i) indicates that reaction A can be written as the
sum of reactions B and C.
(b)
H Z = H X + H Y . Diagram (ii) indicates that reaction Z can be written as
the sum of reactions X and Y.
(c)
Hess’s law states that the enthalpy change for a net reaction is the sum of the
enthalpy changes of the component steps, regardless of whether the reaction
actually occurs via this path.
(d)
No. The enthalpy relationships are true because enthalpy is a state function,
independent of path. Work is not a state function.
Because mass must be conserved in the reaction A  B, the component elements of A
and B must be the same. Further, if  H of > 0 for both A and B, the energies of both A
and B are above the energies of their component elements on the energy diagram.
(a)
The bold arrow shows the reaction as written;
combination of the two thin arrows shows an
alternate route from A to B.
(b)
 H orxn   H of B   H of A .If the overall reaction is exothermic, the sign of H is (–)
and  H of A   H of B . This means that the enthalpy of A is the highest energy
level on the diagram. This is the situation pictured in the diagram above, but
nothing in the given information requires this arrangement. If the reaction is
endothermic,  H of B   H of A and the enthalpy of B would be the highest energy
level on the diagram.
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5 Thermochemistry
Solutions to Exercises
The Nature of Chemical Energy (Section 5.1)
5.13
Analyze/Plan. Use Equation 5.2 to calculate electrostatic potential energy, Ee1. The
distances between the particles are given in the exercise. The charge of an electron and a
proton are given in Section 5.1.
Solve.
(a)
el  κ
Q1  Q1
J-m (1.60  1019 C)  (1.60  1019 C)
 8.99  109 2
  1.0  1018 J
d
C
230  1012 m
(b)
el  κ
Q1  Q1
J-m (1.60  1019 C)  (1.60  1019 C)
 8.99  109 2
  2.3  1019 J
d
C
1.0  109 m
The change in potential energy is 7.7  10–19 J
5.14
(c)
The two objects carry opposite charges; they attract each other. As they move
farther apart, their energy increases.
(a)
el  κ
Q1  Q1
J-m (1.60  1019 C)  (1.60  1019 C)
 8.99  109 2
 5.0  1019 J
d
C
460  1012 m
(b)
el  κ
Q1  Q1
J-m (1.60  1019 C)  (1.60  1019 C)
 8.99  109 2
 2.3  1019 J
d
C
1.0  109 m
The change in potential energy is 2.7  10–19 J
(c)
5.15
The two objects carry like charges; they repulse each other. As they move farther
apart, their energy decreases.
Analyze/Plan. Use the equation for electrostatic attractive force and the distance
between particles given in the exercise. Find the charges of a proton and an
electron in Section 5.1. Solve.
Q1  Q1
J-m ( 1.60  1019 C)  (1.60  1019 C)
 8.99  109 2
  4.4  109 N
9
2
d
C
(0.23  10 m)
(a)
el  κ
(b)
m e = 9.11  10–31 kg; m p = 1.67  10–27 kg
Fg 
m1  m1
d2
 6.674  1011
N-m 2 (9.11  1031 kg)  (1.673  1027 kg)
kg 2
(0.23  109 m) 2
 1.9  1048 N
5.16
(c)
The electrostatic force is 2.3  1039 times larger than the gravitational force. Both
forces are attractive, that is, they point in the same direction.
(a)
Fel  κ
(b)
Fg  G
Q1  Q 2
d
2
m1  m 2
d2
 8.99  109
J-m ( 1.60  1019 C)  (1.60  1019 C)

= 4.1  108 N
2
12
2
C
(75  10
m)
 6.674  1011
N-m 2 (9.11  1031 kg)  (9.11  1031 kg)
kg 2
(75  1012 m) 2
= 9.8  1051 N
(c)
5.17
The repulsive electrostatic force is much larger than the attractive gravitational
force, so the electrons will be repelled from one another.
Analyze/Plan. We must find the work required to completely separate two oppositely
charged particles. Work is the energy required to move an object against a force
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5 Thermochemistry
Solutions to Exercises
(Section 1.4). At infinite separation, the electrostatic potential energy of the pair of ions is
zero. The magnitude of the work required is equal to the electrostatic potential energy,
E e l, of the pair of ions. Use Equation 5.2 to calculate E e l and work. The charges of the
ions and the distance between them are given in the exercise. Solve.
el 
el 
 Q 1Q 2
d
; d = 0.50 nm; Q C l = –1.6  10–19 C; Q N a = 1.6  10–19 C
8.99  109 Jm
C  0.50 nm
2

1nm
1 109 m
  1.6  1019 C  1.6  1019 C 
 4.603  1019   4.6  1019 J
The sign of E e l is negative, so the work required to separate the ions is 4.6  10–19 J.
5.18
el 
el 
 Q 1Q 2
d
; d = 0.35 nm; Q O = –3.2  10–19 C; Q M g = 3.2  10–19 C
8.99  109 J-m
C  0.35 nm
2

1nm
1 109 m
  3.2  1019 C  3.2  1019 C 
 2.630  1018  2.6  1018 J
The sign of E e l is negative, so the work required to separate the ions is 2.6  10–18 J.
5. 19
5.20
(a)
Gravity; work is done because the force of gravity is opposed and the book is
lifted.
(b)
Gas spring force; work is done to compress the gas in the pump.
(a)
Electrostatic attraction; no work is done because the direction of this force is
always perpendicular to the movement of the electron.
(b)
Magnetic attraction; work is done because the nail is moved a distance in
attraction to the magnetic force.
The First Law of Thermodynamics (Section 5.2)
5.21
5.22
5.23
(a)
Matter cannot leave a closed system. Energy in the form of heat or work can be
transferred between a closed system and the surroundings.
(b)
Neither matter nor energy can leave or enter an isolated system.
(c)
Any part of the universe not part of the system is called the surroundings.
(a)
The liquid is an open system because it exchanges both matter and energy with the
surroundings. Matter exchange occurs when solution flows into and out of the
apparatus. The apparatus is not insulated, so energy exchange also occurs. Closed
systems exchange energy but not matter, whereas isolated systems exchange neither.
(b)
If the inlet and outlet are closed, the system can exchange energy but not matter
with the surroundings; it becomes a closed system.
(a)
According to the first law of thermodynamics, energy is conserved.
(b)
The total internal energy (E) of a system is the sum of all the kinetic and potential
energies of the system components.
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5 Thermochemistry
5.24
5.25
Solutions to Exercises
(c)
The internal energy of a closed system (where no matter exchange with
surroundings occurs) increases when work is done on the system by the
surroundings and/or when heat is transferred to the system from the surroundings
(the system is heated).
(a)
E = q + w
(b)
The quantities q and w are negative when the system loses heat to the
surroundings (it cools) or does work on the surroundings.
Analyze. Given: heat and work. Find: magnitude and sign of E.
Plan. In each case, evaluate q and w in the expression E = q + w. For an exothermic
process, q is negative; for an endothermic process, q is positive. Solve.
5.26
5.27
(a)
q = 0.763 kJ, w = –840 J = –0.840 kJ. E = 0.763 kJ – 0.840 kJ = – 0.077 kJ. The
process is endothermic.
(b)
q is negative because the system releases heat, and w is positive because work is
done on the system. E = –66.1 kJ + 44.0 kJ = –22.1 kJ. The process is exothermic.
In each case, evaluate q and w in the expression E = q + w. For an exothermic process,
q is negative; for an endothermic process, q is positive.
(a)
q is negative and w is positive. E = –0.655 kJ + 0.382 kJ = –0.273 kJ. The process
is exothermic.
(b)
q is positive and w is essentially zero. E = 322 J. The process is endothermic.
Analyze. How do the different physical situations (cases) affect the changes to heat and
work of the system upon addition of 100 J of energy?
Plan. Use the definitions of heat and work and the First Law to answer the questions.
Solve. If the piston is allowed to move, case (1), the heated gas will expand and push
the piston up, doing work on the surroundings. If the piston is fixed, case (2), most of
the electrical energy will be manifested as an increase in heat of the system.
5.28
(a)
Because little or no work is done by the system in case (2), the gas will absorb
most of the energy as heat; the case (2) gas will have the higher temperature.
(b)
In Case 1, w is negative because work is done on the surroundings by expansion.
Because the transfer of electrical energy is never completely efficient and some
energy will be transferred as heat, q is positive. In Case 2, w is zero because no
work (expansion) is done. The value of q is positive because all energy is
transferred as heat.
(c)
E is greater for case (2) because the entire 100 J increases the internal energy of
the system, rather than a part of the energy doing work on the surroundings.
κ Q 1Q 2
For two oppositely charged particles, the sign of E e l is negative; the closer
r
the particles, the greater the magnitude of E e l.
el 
(a)
The potential energy of oppositely charged spheres increases (becomes less
negative) as the particles are separated (r increases).
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5 Thermochemistry
5.29
5.30
Solutions to Exercises
(b)
E for the process is positive; the internal energy of the system increases as the
oppositely charged particles are separated.
(c)
Work is done on the system to separate the particles so w is positive. Mechanical
separation of macroscopic charged spheres involves no heat transfer. Work alone
accounts for the change in energy of the system.
(a)
A state function is a property of a system that depends only on the physical state
(pressure, temperature, etc.) of the system, not on the route used by the system to
get to the current state.
(b)
Internal energy and enthalpy are state functions; heat is not a state function.
(c)
Volume is a state function. The volume of a system depends only on conditions
(pressure, temperature, amount of substance), not the route or method used to
establish that volume.
(a)
Independent. Potential energy is a state function.
(b)
Dependent. Some of the energy released could be employed in performing work,
as is done in the body when sugar is metabolized; heat is not a state function.
(c)
Dependent. The work accomplished depends on whether the gasoline is used in
an engine, burned in an open flame, or in some other manner. Work is not a state
function.
Enthalpy (Sections 5.3 and 5.4)
5.31
Analyze. Given, P = 101.3 kPa, ΔV = +2.0 L. Find work involved, in J.
Plan. This change is P—V work done at constant P. w =−PΔV. 1 L‐kPa = 1 J
Solve. w = −101.3 kPa (2.0 L) = −202.6 L‐kPa; = −202.6 J = −203 J
The negative sign indicates that work is done by the system on the surroundings.
5.32
The change in volume is negative: ∆V = Vfinal − Vinitial = 11.2 L − 33.6 L = −22.4 L,
so the p−V work has a positive sign
As the pressure remains constant during the change of the volume (isobaric process),
work can be calculated using w = −P∆V
w = −P ∆V= −90.5 kPa × (−22.4 L) = 2026 L-kPa = 2026 J
5.33
(a)
As ∆E = q + w, measuring ∆E generally requires a means to measure both q and
w, while, at constant pressure, the measured heat qp is identical to the change in
enthalpy ∆H.
(b)
E is a property of a thermodynamic system; it only depends on its conditions,
e.g., its temperature and pressure. Heat q is not a property of a system, q
describes energy transferred to or from the system. We can equate change in
internal energy, ∆E, with heat, only for the specific conditions of constant volume
and no work.
(c)
If ∆H is negative, the enthalpy of the system decreases, and the process is
exothermic.
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5 Thermochemistry
5.34
5.35
Solutions to Exercises
(a)
When a process occurs under constant external pressure and only P–V work
occurs, the enthalpy change (H) equals the amount of heat transferred. H = q p .
(b)
H = q p . If the system releases heat, q and H are negative, and the enthalpy of
the system decreases.
(c)
If H = 0, q p = 0 and E = w .
(a)
w = –P∆V. If we assume the gases to be ideal, ∆V = RT∆n . During the reaction, 3
moles of gas are converted to liquid water, so ∆n = –3.
w = –P∆V = –RT∆n = –8.314 J/mol k × 298 K × (–3) = 7.4 kJ
(b)
∆H = –483.6 kJ
∆E = ∆H – P∆V = ∆H + w = –483.6 kJ + 7.4 kJ = –476.2 kJ
5.36
5.37
(a)
At constant volume (V = 0), E = q v .
(b)
E will be larger than H.
(c)
According to the definition of enthalpy, H = E + PV, so H = E + (PV). For an
ideal gas at constant temperature and volume, PV = VP = RTn. For this
reaction, there are 2 mol of gaseous product and 3 mol of gaseous reactants, so n
= –1. Thus VP or (PV) is negative. Because H = E + (PV), the negative
(PV) term means that E is larger or less negative than H.
Analyze/Plan. q = 824 J = 0.824 kJ (heat is absorbed by the system), w = 0.65 kJ (work is
done on the system). Solve.
E = q + w = 0.824 kJ + 0.65 kJ = 1.47 kJ. H = q = 0.824 kJ (at constant pressure).
Check. The reaction is endothermic.
5.38
The gas is the system. If 0.49 kJ of heat is added, q = +0.49 kJ. Work done by the system
decreases the overall energy of the system, so w = –214 J = –0.214 kJ .
E = q + w = 0.49 kJ – 0.214 kJ = 0.276 kJ. H = q = 0.49 kJ (at constant pressure).
5.39
(a)
CH 4 (g) + 2 O 2 (g)  CO2 (g) + 2 H2 O(l)
(b)
Analyze. How are reactants and products
arranged on an enthalpy diagram?
H = –890 kJ
Plan. The substances (reactants and products,
collectively) with higher enthalpy are shown on
the upper level, and those with lower enthalpy
are shown on the lower level.
Solve. For this reaction, H is negative, so the products have lower enthalpy and
are shown on the lower level; reactants are on the upper level. The arrow points
in the direction of reactants to products and is labeled with the value of H.
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5 Thermochemistry
5.40
5.41
5.42
5.43
Solutions to Exercises
(a)
2 NaHCO3 (s)  Na2CO3 (s) + H2O (l) + CO2 (g)
(b)
H = +85 kJ
Analyze/Plan. Consider ∆H for the exothermic reaction as written. Solve.
(a)
ΔHrxn0 = (0 kJ/mol) – 2(217.94 kJ/mol) = –435.88 kJ/(mol )
(b)
Because H is negative, the reactants, 2 H (g) has the higher enthalpy.
Plan. Consider the sign of an enthalpy change that would convert one of the substances
into another.
(a)
I2 (s)  I2 (g). This change is sublimation, which is endothermic; I2 (g) has the
higher enthalpy.
(b)
I2 (g)  2 I(g). Breaking the I–I bond requires energy, so the process is
endothermic. Two moles of I atoms have higher enthalpy.
(c)
2 H I (g)  I 2 (g) + H2 (g). Decomposing HI into its elements requires energy and
is endothermic. The elements have the higher enthalpy.
(d)
H 2 (g) at 100 C  H 2 (g) at 300 C. An increase in temperature of the sample
requires that heat is added to the system, thus it is endothermic. H 2 (g) at 300 C
has the higher enthalpy.
Analyze/Plan. Follow the strategy in Sample Exercise 5.4.
(a)
Exothermic (H is negative)
(b)
3.55 g M g 
Solve.
1m olM g
 1204 kJ

  87.9 kJ heat transferred
24.305 g M g 2 m olM g
Check. The units of kJ are correct for heat. The negative sign indicates heat is
evolved.
(c)
 234 kJ
2 m olM gO 40.30 g M gO

 15.7 g M gO produced
1m olM g
 1204 kJ
Check. Units are correct for mass. (200  2  40/1200)  (16,000/1200) > 10 g
(d)
2 MgO(s)  2 Mg(s) + O 2 (g)
H = +1204 kJ
This is the reverse of the reaction given above, so the sign of H is reversed.
40.3 g M gO 
1m olM gO
1204 kJ

  602 kJ heat absorbed
40.30 g M gO 2 m olM gO
Check. 40.3 g MgO is just 1 mol MgO, so the calculated value is the heat absorbed
per mol of MgO, 1204 kJ/2 mol MgO = 602 kJ.
5.44
(a)
The sign of H is positive, so the reaction is endothermic.
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5 Thermochemistry
1 m olC H 3O H
252.8 kJ

 94.7 kJ heat absorbed
32.04 g C H 3O H 2 m olC H 3O H
(b)
24.0 g C H 3O H 
(c)
82.1 kJ
(d)
The sign of H is reversed for the reverse reaction: H = 252.8 kJ
2 m olC H 4 16.04 g C H 4

 10.4 g C H 4 produced
252.8 kJ
1 m olC H 4
38.5 g C H 4 
5.45
Solutions to Exercises
1 m olC H 4
 252.8 kJ

  303 kJ heat released
16.04 g C H 4 2 m olC H 4
Analyze. Given: balanced thermochemical equation, various quantities of substances
and/or enthalpy. Plan. Enthalpy is an extensive property; it is “stoichiometric.” Use the
mole ratios implicit in the balanced thermochemical equation to solve for the desired
quantity. Use molar masses to change mass to moles and vice versa where appropriate.
Solve.
 65.5 kJ
  29.5 kJ
0.450 m olA gC l
(a)
1m olA gC l
Check. Units are correct; sign indicates heat evolved.
(b)
9.00 g A gC l
 65.5 kJ
1m olA gC l

  4.11kJ
143.3 g A gC l 1m olA gC l
Check. Units correct; sign indicates heat evolved.
(c)
9.25  104 m olA gC l
 65.5 kJ
 0.0606 kJ 60.6 J
1m olA gC l
Check. Units correct; sign of H reversed; sign indicates heat is absorbed during
the reverse reaction.
5.46
5.47
 89.4 kJ
  40.53   40.5 kJ
3 m olO 2
(a)
1.36 m olO 2 
(b)
10.4 g K C l
(c)
Because the sign of H is reversed for the reverse reaction, it seems reasonable
that other characteristics would be reversed, as well. If the forward reaction
proceeds spontaneously, the reverse reaction is probably not spontaneous. Also,
we know from experience that KCl(s) does not spontaneously react with
atmospheric O 2 (g), even at elevated temperature.
1 m olK C l
 89.4 kJ

 6.2358   6.24 kJ
74.55 g K C l 2 m olK C l
Analyze. Given: balanced thermochemical equation. Plan. Follow the guidelines given in
Section 5.4 for evaluating thermochemical equations. Solve.
(a)
When a chemical equation is reversed, the sign of H is reversed.
CO 2 (g) + 2 H 2 O(l)  CH 3 OH(l) + 3/2 O 2 (g)
(b)
H = +726.5 kJ
Enthalpy is extensive. If the coefficients in the chemical equation are multiplied
by 2 to obtain all integer coefficients, the enthalpy change is also multiplied by 2.
2 CH 3 OH(l) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(l)
H = 2(–726.5) kJ = –1453 kJ
(c)
The exothermic forward reaction is more likely to be thermodynamically favored.
(d)
Decrease. Vaporization (liquid  gas) is endothermic. If the product were
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5 Thermochemistry
Solutions to Exercises
H 2 O(g), the reaction would be more endothermic and would have a smaller
negative H. (Depending on temperature, the enthalpy of vaporization for 2 mol
H 2 O is about +88 kJ, not large enough to cause the overall reaction to be
endothermic.)
5.48
(a)
3 C 2 H 2 (g)  C 6 H 6 (l)
H = –630 kJ
(b)
C 6 H 6 (l)  3 C 2 H 2 (g)
H = +630 kJ
H for the formation of 3 mol of acetylene is 630 kJ. H for the formation of 1 mol
of C2H2 is then 630 kJ/3 = 210 kJ.
(c)
The exothermic reverse reaction is more likely to be thermodynamically favored.
If the reactant is in the higher enthalpy gas phase, the overall H for the reaction
decreases.
Calorimetry (Section 5.5)
The specific heat of water to four significant figures, 4.184 J/g-K, will be used in many
of the following exercises; temperature units of K and C will be used interchangeably.
5.49
(a)
Heat capacity is the amount of heat in J required to raise the temperature of an
object or a certain amount of substance 1 ° C or 1 K.
q
m  T
Molar heat capacity is the heat capacity of one mole of substance.
q
Cm 
n  T
Dividing these two equations leads to,
Cs 
q
Cm
m
T
 n
  MM
q
Cs
n
m T
(b)
(c)
5.50
So, the quotient of molar heat capacity and specific heat is simply the molar
mass.
j
g
Cm  Cs  MM  0.9
 26.98
 24.3 J/mol-k
g-k
mol
To calculate heat capacity from specific heat, the mass of the particular piece of
aluminum component must be known.
Analyze. Both objects are heated to 100 C. The two hot objects are placed in the same
amount of cold water at the same temperature. Object A raises the water temperature
more than object B. Plan. Apply the definition of heat capacity to heating the water and
heating the objects to determine which object has the greater heat capacity. Solve.
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5.51
Solutions to Exercises
(a)
Both beakers of water contain the same mass of water, so they both have the
same heat capacity. Object A raises the temperature of its water more than object
B, so more heat was transferred from object A than from object B. Because both
objects were heated to the same temperature initially, object A must have
absorbed more heat to reach the 100 C temperature. The greater the heat
capacity of an object, the greater the heat required to produce a given rise in
temperature. Thus, object A has the greater heat capacity.
(b)
Because no information about the masses of the objects is given, we cannot
compare or determine the specific heats of the objects.
Plan. Manipulate the definition of specific heat to solve for the desired quantity, paying
close attention to units. Cs = q/(m  T). Solve.
(a)
4.184
J
 1 g  1 K = 4.184 J
g -K
(b)
4.184
J
 18.02 g  1 K = 75.4 J
g -K
(c)
C = Cs × m = 4.184
J
 370 g = 1548 J/K
g -K
T  (46.2 C  24.6 o C)  21.6 o C = 21.6 K
(d)
q = Cs × m × T= 4.184
5.52
5.53
J
g
 5.00 kg 1000  (46.2  24.6)K = 452 kJ
g -K
kg
(a)
In Table 5.2, Hg(l) has the smallest specific heat, so it will require the smallest
amount of energy to heat 50.0 g of the substance 10 K.
(b)
50.0 g H g(l) 10 K 
0.14 J
 70 J
g-K
Analyze/Plan. Follow the logic in Sample Exercise 5.5.
(a)
(b)
Solve.
2.22 J
 (25.0 o C  10.0 o C ) 2.66 103 J (or 2.66 kJ)
g-K
Plan. Calculate the molar heat capacity of octane and compare it with the molar
heat capacity of water, 75.40 J/mol-oC, as calculated in Exercise 5.51(b). Solve.
80.0 g C 8H 18 
2.22 J 114.2 g C 8H 18 253.58 J 254 J



g-K
1m olC 8H 18
m ol-K
m ol-K
The molar heat capacity of C8H18(l), 254 J/mol-K, is greater than that of H2O(l), so
it will require more heat to increase the temperature of octane than to increase the
temperature of water.
5.54
322 J
J
0.13 J

 0.1288 
1g-oC 100.0 g  (50 o C  25 oC )
1 g-oC
(a)
specific heat
(b)
In general, the greater the heat capacity, the more heat is required to raise the
temperature of 1 gram of substance 1 oC. The specific heat of gold is 0.13 J/g-oC,
whereas that of iron is 0.45 J/g-oC (Table 5.2). For gold and iron blocks with equal
mass, same initial temperature and same amount of heat added, the one with the
lower specific heat, gold, will require less heat per oC and have the higher final
temperature.
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5 Thermochemistry
(c)
5.55
Solutions to Exercises
0.1288 J 196.97 g A u
25 J

 25.37 
o
1m olA u
1 g- C
m ol-oC
Analyze. Because the temperature of the water increases, the dissolving process is
exothermic and the sign of H is negative. The heat lost by the NaOH(s) dissolving
equals the heat gained by the solution.
Plan/Solve. Calculate the heat gained by the solution. The temperature change is
37.8 – 21.6 = 16.2 C. The total mass of solution is (100.0 g H2O + 6.50 g NaOH) = 106.5 g.
106.5 g solution 
4.184 J
1 kJ
 16.2 o C 
 7.2187  7.22 kJ
1000 J
1 g-oC
This is the amount of heat lost when 6.50 g of NaOH dissolves.
The heat loss per mole NaOH is
40.00 g N aO H
7.2187 kJ

  44.4 kJ/m ol
6.50 g N aO H
1 m olN aO H
H  qp   44.4 kJ/m olN aO H
Check. (–7/7  40)  –40 kJ; the units and sign are correct.
5.56
(a)
Follow the logic in Solution 5.55. The total mass of the solution is (60.0 g
H 2 O + 4.25 g NH 4 NO 3 ) = 64.25 = 64.3 g. The temperature change of the solution
is 22.0 – 16.9 = –5.1 C. The heat lost by the surroundings is
64.25 g solution 
4.184 J
1 kJ
 5.1 oC 
  1.371  1.4 kJ
1 g-°C
1000 J
That is, 1.4 kJ is absorbed when 4.25 g NH 4 NO 3 (s) dissolves.
80.04 g N H 4N O 3
1.371 kJ

  25.82   26 kJ/m olN H 4N O 3
4.25 N H 4N O 3
1 m olN H 4N O 3
5.57
(b)
This process is endothermic because the temperature of the surroundings
decreases, indicating that heat is absorbed by the system.
(a)
C 6 H 4 O 2 (s) + 6 O 2 (g)  2 H2 O(l) + 6 CO2 (g)
(b)
qbomb = − qrxn; ∆T = 29.49°C – 25.00°C = 4.49°C
q bomb 
8.500 kJ
 4.49o C  38.17 kJ
1o C
At constant volume, q v = E. E and H are very similar.
H rxn  E rxn   q rxn  q bomb 
 H rxn 
 38.17 kJ
 25.44 kJ/g C6 H 4 O 2
1.50 g C6 H 4 O 2
 25.44 kJ 108.1 g C6 H 4 O 2

1 g C6 H 4 O 2
1 mol C6 H 4 O 2
  2, 750 kJ/mol  2.750  103 kJ/mol C6 H 4 O 2
5.58
(a)
C 6 H 5 OH(s) + 7 O 2 (g)  6 CO 2 (g) + 3 H 2 O(l)
(b)
qbomb = –qrxn; T = 26.37 C – 21.36 C = 5.01 C
qbom b 
11.66 kJ
 5.01 o C  58.417  58.4 kJ
1 oC
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5 Thermochemistry
Solutions to Exercises
At constant volume, q v = E. E and H are very similar.
 H rxn   Erxn  qrxn  qbom b 
 H rxn 
 58.417 kJ
  32.454   32.5 kJ/g C 6H 5O H
1.800 g C 6H 5O H
32.454 kJ 94.11 g C 6H 5O H
 3.054 103 kJ


1 g C 6H 5O H
1 m olC 6H 5O H
m olC 6H 5O H
= –3.05  10 3 kJ/mol C 6 H 5 OH
5.59
Analyze. Given: specific heat and mass of glucose, T for calorimeter. Find: heat
capacity, C, of calorimeter. Plan. All heat from the combustion raises the temperature of
the calorimeter. Calculate heat from combustion of glucose, divide by T for
calorimeter to get kJ/C. T = 24.72 C – 20.94 C = 3.78 C Solve.
15.57 kJ
1

 14.42  14.4 kJ/ o C
1 g glucose 3.78 o C
(a)
C total  3.500 g glucose 
(b)
Qualitatively, assuming the same exact initial conditions in the calorimeter, twice
as much glucose produces twice as much heat, which raises the calorimeter
temperature by twice as many C. Quantitatively,
7.000 g glucose 
1 oC
15.57 kJ

 7.56 o C
1 g glucose 14.42 kJ
Check. Units are correct. T is twice as large as in part (a). The result has 3 sig figs,
because the heat capacity of the calorimeter is known to 3 sig figs.
5.60
26.38 kJ
1

 8.74055  8.74 kJ/oC
1 g C 6H 5C O O H 8.33 o C
(a)
C  2.760 g C 6H 5C O O H 
(b)
8.74055 kJ
1
 4.95 oC 
 30.046  30.0 kJ/g sam ple
o
1.440 g sam ple
C
(c)
If water is lost from the calorimeter, the heat capacity of the calorimeter decreases.
Hess’s Law (Section 5.6)
5.61
Yes, because internal energy is a state function. Hess’s Law works for any state
function.
5.62
(a)
Analyze/Plan. Arrange the reactions so that in the overall sum, B appears in both
reactants and products and can be canceled. This is a general technique for using
Hess’s Law. Solve.
AB
 H I   60 kJ
BC
AC
 H II   90 kJ
H III   30 kJ
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5 Thermochemistry
Solutions to Exercises
(b)
Check. The process of A forming C can be described as A forming B and B
forming C.
5.63
Analyze/Plan. Follow the logic in Sample Exercise 5.8. Manipulate the equations so that
“unwanted” substances can be canceled from reactants and products. Adjust the
corresponding sign and magnitude of H. Solve.
P4O 6 (s)  P4 (s)+ 3O 2(g)
 H  1640.1 kJ
P4 (s)+ 5 O 2 (g)  P4O 10 (s)
 H  2940.1kJ
P4O 6 (s)+ 2 O 2 (g)  P4O 10 (s)
 H = 1300.0 kJ
Check. We have obtained the desired reaction.
5.64
5.65
2 C (s)  O 2 (g)  4 H 2 (g )  2 C H 3O H (g )
H  402.4 kJ
2 C O (g)  O 2 (g)  2 C(s)
H  221.0 kJ
2 C O (g)  4 H 2 (g)  2 C H 3O H (g )
H  181.4 kJ
C O (g )  2 H 2 (g )  C H 3O H (g )
H  (181.4)/ 2   90.7 kJ
Analyze/Plan. Follow the logic in Sample Exercise 5.9. Manipulate the equations so that
“unwanted” substances can be canceled from reactants and products. Adjust the
corresponding sign and magnitude of H. Solve.
C 2H 4 (g)

2 H 2 (g) 2 C (s)
H   52.3 kJ
2 C (s) 4 F2 (g)

2 C F4 (g)
 H  2( 680 kJ)
2 H 2 (g) 2 F2 (g)

4 H F(g)
H  2( 537 kJ)
C 2H 4 (g) 6 F2 (g)

2 C F4 (g) 4 H F(g)
H   2.49 103 kJ
Check. We have obtained the desired reaction.
5.66
N 2 (g) 
N 2O (g)  N 2 (g)  1/2 O 2 (g)
H  1/2 ( 163.2 kJ)
N O 2 (g)  N O (g)  1/2 O 2 (g)
H  1/2(113.1 kJ)
 2 N O (g)
H  180.7 kJ
N 2O (g)  N O 2 (g)  3 N O (g)
H  155.7 kJ
O 2 (g)
Enthalpies of Formation (Section 5.7)
5.67
(a)
Standard conditions for enthalpy changes are usually P = 1 atm and T = 298 K. For
the purpose of comparison, standard enthalpy changes, Hº , are tabulated for
reactions at these conditions.
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5 Thermochemistry
5.68
5.69
5.70
5.71
Solutions to Exercises
(b)
Enthalpy of formation, H f , is the enthalpy change that occurs when a compound
is formed from its component elements.
(c)
Standard enthalpy of formation,  H of , is the enthalpy change that accompanies
formation of 1 mole of a substance from elements in their standard states.
(a)
The standard enthalpy of formation for any element in its standard state is
arbitrarily defined as zero, regardless of its chemical reactivity. The standard
enthalpy of formation is no measure of stability.
(b)
10 C(s) + 4 H 2 (g)  C 1 0 H 8(s)
(a)
1/2 N 2 (g) + O 2 (g)  NO 2 (g)
 H of 
(b)
S (s) + 3/2 O 2 (g)  SO 3 (g)
 H of   395.2 kJ
(c)
Na(s) + 1/2 Br 2 (l)  NaBr (s)
 H of   361.4 kJ
(d)
Pb(s) N 2(g) 3 O 2 (g)  Pb(N O 3 )2 (s)
 H of   451.9 kJ
(a)
C(s) + 2H 2 (g) + 1/2 O 2 (g)  CH 3 OH(l)
 H of   238.6 kJ/mol
(b)
Ca(s) + S (s) + 2 O2 (g)  CaSO4 (s)
 H of   1, 434.0 kJ /mol
(c)
1/2 N2 (g) + 1/2 O2 (g)  NO (g)
 H of   90.37 kJ /mol
(d)
P4 (s) + 3 O2 (g)  P4 O6 (s)
 H of   1, 640.1 kJ /mol
33.84 kJ
o
  n H of (products)  n H of (reactants).Be careful with coefficients,
Plan.  H rxn
states, and signs. Solve.
o
 H rxn
 H of A l2O 3 (s) 2 H of Fe(s) H of Fe2O 3 (s) 2 H of A l(s)
o
H rxn
 ( 1669.8 kJ) 2(0) ( 822.16 kJ) 2(0)  847.6 kJ
5.72
Use heats of formation to calculate Hº for the combustion of butane.
C 3H 8 (g)  5 O 2 (g) 3 C O 2 (g) 4 H 2O (l)
o
 H rxn
 3 H of C O 2 (g) 4 H of H 2O (l) H of C 3H 8 (g) 5 H of O 2 (g)
o
 H rxn
 3( 393.5 kJ) 4( 285.83 kJ) ( 103.85 kJ) 5(0) 2219.97   2220.0 kJ/m olC 3H 8
10.00 g C 3H 8 
5.73
1 m olC 3H 8
 2219.97 kJ

  503.4 kJ
44.096 g C 3H 8 1 m olC 3H 8
o
  n H of (products)  n H of (reactants). Be careful with coefficients,
Plan.  H rxn
states, and signs. Solve.
(a)
o
 H rxn
 2  H of SO 3 (g) 2  H of SO 2 (g)  H of O 2 (g)
= 2(–395.2 kJ) – 2(–296.9 kJ) – 0 = –196.6 kJ
(b)
o
 H rxn
  H of M gO (s) H of H 2O (l) H of M g(O H )2 (s)
= –601.8 kJ + (–285.83 kJ) – (–924.7 kJ) = 37.1 kJ
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Solutions to Exercises
o
 H rxn
 4 H of H 2O (g) H of N 2 (g) H of N 2O 4 (g) 4 H of H 2 (g)
(c)
= 4(–241.82 kJ) + 0 – (9.66 kJ) – 4(0) = –976.94 kJ
   H  SiO (s) 4 H  H C l(g) H  SiC l (l) 2  H  H O (l)
 H rxn
f
2
f
f
4
f
2
(d)
= –910.9 kJ + 4(–92.30 kJ) – (–640.1 kJ) – 2(–285.83 kJ) = –68.3 kJ
5.74
o
 H rxn
  H fo H 2 O(g)  H fo CaF2 (g)   H fo CaO(s)  2 H fo HF(g)
(a)
kJ  
kJ  
kJ  
kJ 

o
 H rxn
   241.82
    1219.6 mol    635.5 mol   2  268.61 mol 
mol

 
 
 

  288.7 kJ/mol
o
 H rxn
 2  H fo Fe(s)  3  H fo CO(g)   H fo Fe 2 O3 (s)  3 H fo C(s)
(b)
kJ  
kJ   kJ 
 kJ  
o
 H rxn
 2 0
 3   110.5
   822.16
3 0


mol  
mol   mol 
 mol  
 490.7 kJ/mol
o
 H rxn
 H fo N 2 (g)  2 H fo CO 2 (g)  2H fo CO(g)  2  H fo NO(g)
(c)
kJ  
kJ  
kJ 
 kJ 

o
 H rxn
 0
  2  393.5 mol   2   110.5 mol   2  90.37 mol 
mol



 
 

  746.7 kJ/mol
o
 H rxn
 4  H fo NO(g)  6 H fo H 2 O(g)  4 H fo NH 3 (g)  5 H fo O 2 (g)
(d)
kJ  
kJ  
kJ   kJ 

o
 H rxn
 4  90.37
 6  241.82
 4   46.19
5 0


mol  
mol  
mol   mol 

  904.7 kJ/mol
5.75
Analyze. Given: combustion reaction, enthalpy of combustion, enthalpies of formation
for most reactants and products. Find: enthalpy of formation for acetone.
Plan. Rearrange the expression for enthalpy of reaction to calculate the desired enthalpy
of formation. Solve.
 H   3 H  C O (g) 3 H  H O (l) H  C H O (l) 4 H  O (g)
rxn
f
2
f
2
f
3
6
f
 1790 kJ 3(393.5 kJ) 3( 285.83 kJ)  H f C 3H 6O (l) 4(0)
 H  C H O (l) 3( 393.5 kJ) 3( 285.83 kJ) 1790 kJ  248 kJ
f
5.76
3
2
6
   H  C a(O H ) (s)  H  C H (g) 2  H  H O (l)  H  C aC (s)
 H rxn
f
2
f
2 2
f
2
f
2
 127.2 kJ  986.2 kJ 226.77 kJ 2( 285.83 kJ) H f C aC 2 (s)
 H  for C aC (s)  60.6 kJ
f
5.77
2
(a)
C 8 H 1 8(l) + 25/2 O 2 (g)  8 CO 2 (g) + 9 H 2 O(g)
H  = –5064.9 kJ
(b)
Plan. Follow the logic in Solution 5.75 and 5.76.
Solve.
  8  H  C O (g) 9  H  H O (g)  H  C H (l) 25/2 H  O (g)
 H rxn
f
2
f
2
f
8 18
f
2

 5064.9 kJ 8( 393.5 kJ) 9( 241.82 kJ) H C H (l) 25/2(0)
H  C H
f
5.78
(a)
8
f
8
18
18 (l) 8( 393.5 kJ) 9( 241.82 kJ) 5064.9 kJ  259.5 kJ
C 4 H 1 0O(l) + 6 O 2 (g)  4 CO 2 (g) + 5 H 2 O(l)
H  = 2723.7 kJ
127
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5 Thermochemistry
(b)
Solutions to Exercises
  4  H  C O (g) 5  H  H O (l)  H  C H O (l) 6 H  O (g)
 H rxn
f
2
f
2
f
4 10
f
2
 2723.7  4( 393.5 kJ) 5( 285.83 kJ)  H f C 4H 10O (l) 6(0)
 H  C H O (l) 4( 393.5 kJ) 5(285.83 kJ) 2723.7 kJ  279.45   279.5 kJ
f
5.79
4
10
(a)
C 2 H 5 OH(l) + 3 O 2 (g)  2 CO 2 (g) + 3 H 2 O(g)
(b)
  2  H  C O (g) 3  H  H O (g) H  C H O H (l) 3 H  O (g)
 H rxn
f
2
f
2
f
2 5
f
2
= 2(393.5 kJ) + 3(241.82 kJ)  (277.7 kJ)  3(0) = 1234.76 = 1234.8 kJ
(c)
Plan. The enthalpy of combustion of ethanol [from part (b)] is 1234.8 kJ/mol.
Change mol to mass using molar mass, then mass to volume using density.
Solve.
 1234.76 kJ 1m olC 2H 5O H 0.789 g 1000 m L
×
×
×
=  21,147 =  2.11× 104 kJ/ L
m olC 2H 5O H
46.06844 g
mL
L
Check. (1200/50)  25; 25  800  20,000
(d)
Plan. The enthalpy of combustion corresponds to any of the molar amounts in the
equation as written. Production of 1234.76 kJ also produces 2 mol CO2. Use this
relationship to calculate mass CO2/kJ.
2 m olC O 2 44.0095 g C O 2
×
= 0.071284 g C O 2 / kJem itted
 1234.76 kJ
m ol
Check. The negative sign associated with enthalpy indicates that energy is
emitted.
5.80
(a)
CH 3 OH(l) + 3/2 O 2 (g)  CO 2 (g) + 2 H 2 O(g)
(b)
   H  C O (g) 2  H  H O (g) H  C H O H (l) 3/ 2 H  O (g)
 H rxn
f
2
f
2
f
3
f
2
= 393.5 kJ + 2(241.82 kJ) (238.6 kJ)  3/2(0) = 638.54 = 638.5 kJ
(c)
 638.54 kJ 1m olC H 3O H 0.791g 1000 m L
 1.58  104 kJ/L produced



m olC H 3O H
32.04 g
mL
L
(d)
1m olC O 2 44.0095 g C O 2
×
= 0.06892 g C O 2/kJem itted
 638.54 kJ
m ol
Bond Enthalpies (Section 5.8)
5.81
(a)
+ H; energy must be supplied to separate oppositely charged ions.
(b)
– H; energy is released when a chemical bond is formed.
(c)
+ H; energy must be supplied to separate a negatively charged electron from a
neutral atom.
(d)
+ H; energy must be supplied to melt a solid.
128
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5 Thermochemistry
5.82
5.83
Solutions to Exercises
(a)
– H; the reactants have four N–O bonds (some of them multiple bonds) while
the products have these same N–O bonds plus an N–N bond. Overall the reaction
involves formation of a new chemical bond and enthalpy decreases.
(b)
– H; energy is released when a chemical bond is formed.
(c)
– H; energy is released when oppositely charged ions form ionic bonds.
(d)
+ H; energy must be supplied to break a chemical bond.
Analyze. Given: structural formulas. Find: enthalpy of reaction.
Plan. Count the number and kinds of bonds that are broken and formed by the reaction.
Use bond enthalpies from Table 5.4 and Equation 5.32 to calculate the overall enthalpy
of reaction, H. Solve.
(a)
H = D(H–H) + D(Br–Br) – 2 D(H–Br)
= 436 kJ + 193 kJ – 2(366 kJ) = –103 kJ
(b)
H = 6 D(C–H) + 2 D(C–O) + 2 D(O–H) + 3 D(O=O)
– 4 D(C=O) – 8 D(O–H)
= 6 D(C–H) + 2 D(C–O) + 3 D(O=O) – 4 D(C=O) – 6 D(O–H)
H = 6(413) + 2(358) + 3(495) – 4(799) – 6(463) = –1295 kJ
5.84
(a)
H = 3 D(C–Br) + D(C–H) + D(Cl–Cl) – 3 D(C–Br) – D(C–Cl) – D(H–Cl)
= D(C–H) + D(Cl–Cl) – D(C–Cl) – D(H–Cl)
H = 413 + 242 – 328 – 431 = –104 kJ
(b)
H = 4 D(C–H) + 2 D(O=O) – 2 D(C=O) – 4 D(O–H)
H = 4(413) + 2(495) – 2(799) – 4(463) = – 808 kJ
5.85
(a)
o
  n H of (products)  nH of (reactants).Be careful with
Plan.  H rxn
coefficients, states, and signs. Solve.
o
 H rxn
 2  H of Br(g)  H of Br2 (g)
= 2(111.8) – 30.71 = 192.9 kJ
This reaction is just the breaking of a Br–Br single bond to form Br atoms;
reactants and products are all in the gas phase. The enthalpy of reaction
represents the bond enthalpy D(Br–Br), 193 kJ.
5.86
(b)
The value of D(Br–Br) in Table 5.4 is 193 kJ, the same as the enthalpy calculated in
part (a). The difference between the two values is zero (to three significant
figures).
(a)
The relevant reaction is N 2 (g)  2 N(g).
o
 H rxn
 2  H of N (g)  H of N 2 (g)
= 2(472.7) – 0 = 945.4 kJ
Our estimate for D(NN) is 945.4 kJ = 945 kJ.
129
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5 Thermochemistry
(b)
Solutions to Exercises
Calculate the overall enthalpy change for the reaction using standard enthalpies
of formation. Use this value for H and bond enthalpies from Table 5.4 to
estimate the enthalpy of the nitrogen-nitrogen bond in N2 H4 .
o
 H rxn
 2  H of N H 3 (g)  H of N 2H 4 (g)  H of H 2 (g)
= 2(–46.19) – 95.40 – 0 = –187.78 = –188 kJ
H = 4 D(N–H) + D(N–N) + D(H–H) – 6 D(N–H)
H = D(N–N) + D(H–H) – 2 D(N–H)
D(N–N) = H – D(H–H) + 2 D(N–H)
D(N–N) = (–188) – (436) + 2(391) = 158 kJ
5.87
(c)
The nitrogen-nitrogen bond in N2 H4 has an enthalpy of 158 kJ; in N2 the enthalpy
is 945 kJ. We are comparing the same pair of bonded atoms, so it is safe to say
that the bond in N2 H4 is weaker than the bond in N2 .
(a)
H = 2 D(H–H) + D(O=O) – 4 D(O–H) = 2(436) + 495 – 4(463) = –485 kJ
(b)
The estimate from part (a) is less negative or larger than the true reaction
enthalpy. When we use bond enthalpies to estimate reaction enthalpies, we
assume all reactants and products are gases. We have estimated the enthalpy
change for production of H 2 O(g). Because condensation, [(g) (l)], is
exothermic, we expect H for production of liquid water to be more negative or
smaller than the value we estimated in part (a).
(c)
o
 H rxn
 2  H of H 2O (l) 2  H of H 2 (g)  H of O 2 (g)
= 2(– 285.83) – 2(0) – 0 = –571.66 = – 572 kJ
As predicted in part (b), the true enthalpy of reaction is more negative than the
result calculated using bond enthalpies.
5.88
(a)
H = D(H–H) + D(Br–Br) – 2 D(H–Br)
kJ  
kJ  
kJ 

 H   436
 193
 2  366
  103 kJ/mol


mol  
mol  
mol 

(b)
To carry out the reaction, not only will the bonds have to be broken, but also
bromine has to be vaporized. So, the true reaction will be less exothermic as
additional energy is required.
(c)
o
 H rxn
 2  H fo HBr(g)   H fo H 2 (g)   H fo Br2 (l)
kJ   kJ   kJ 

 2   36.23
 0
 0
  72.46 kJ/mol
mol   mol   mol 

Foods and Fuels (Section 5.9)
5.89
(a)
Fuel value is the amount of energy produced when 1 gram of a substance (fuel) is
combusted.
(b)
The fuel value of fats is 9 kcal/g and of carbohydrates is 4 kcal/g. Therefore, 5 g
of fat produce 45 kcal, whereas 9 g of carbohydrates produce 36 kcal; 5 g of fat are
a greater energy source.
(c)
These products of metabolism are expelled as waste, H 2 O(l) primarily in urine
and feces, and CO 2 (g) as gas when breathing.
130
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5 Thermochemistry
5.90
Solutions to Exercises
(a)
One gram of fat produces more energy than one gram of carbohydrates when
metabolized.
(b)
For convenience, assume 100 g of chips.
12 g protein 
14 g fat
17 kJ
1C al

 48.76  49 C al
1g protein 4.184 kJ
38 kJ
1 C al

 127.15  130 C al
1 g fat 4.184 kJ
74 g carbohydrates
17 kJ
1 C al

 300.67  301 C al
1 g carbohydrates 4.184 kJ
total Cal = (48.76 + 127.15 + 300.67) = 476.58 = 480 Cal
% C alfrom fat
127.15 C alfat
 100  26.68  27%
476.58 totalC al
(Because the conversion from kJ to Cal was common to all three components, we
would have determined the same percentage by using kJ.)
5.91
38 kJ
17 kJ
 x g protein 
;x  56 g protein
g fat
g protein
(c)
25 g fat
(a)
Plan. Calculate the Cal (kcal) from each nutritional component of the soup, then
sum. Solve.
4.5 g fat 
38 kJ
 171 kJ
1 g fat
42 g carbohydrates 
4.0 g protein 
17 kJ
 714 kJ
1 g carbohydrate
17 kJ
 68 kJ
1 g protein
total energy = 171 kJ + 714 kJ + 68 kJ = 953 kJ
953 kJ 
(b)
5.92
1 kcal
 228 Cal
4.184 kJ
Potassium does not contribute to the calorie content of the food because it is not
metabolized by the body.
Calculate the fuel value in a kJ of one hamburger.
28 g fat 
38 kJ
 1064 kJ
1 g fat
46 g carbohydrate 
25 g protein 
17 kJ
 782 kJ
1 g carbohydrate
17 kJ
 425 kJ
1 g protein
total fuel value = 1064 kJ + 782 kJ + 425 kJ = 2271 kJ
2271 kJ 
1 Cal
 543 Cal
4.184 kJ
131
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5 Thermochemistry
5.93
Solutions to Exercises
Plan. g  mol  kJ  kcal
16.0 g C6 H12 O6 
Solve.
1 mol C6 H12 O6
2812 kJ
1 kcal


 59.7 kcal
180.2 g C6 12 O6 mol C6 12 O6 4.184 kJ
Check. 60 kcal is a reasonable result for most of the food value in an apple.
5.94
6%
1g
 500 mL 
 30 g
100%
mL
kJ
1367 mol
 30 g  890 kJ
g
46.07 mol
890 kJ 
5.95
1 Cal
 213 Cal
4.184 kJ
Plan. Use enthalpies of formation to calculate molar heat (enthalpy) of combustion
using Hess’s Law. Use molar mass to calculate heat of combustion per kg of
hydrocarbon. Solve.
Propyne: C 3 H 4 (g) + 4 O 2 (g)  3 CO 2 (g) + 2 H 2 O(g)
(a)
  3(–393.5 kJ) + 2(–241.82 kJ) – (185.4 kJ) – 4(0) = –1849.5
 H rxn
= –1850 kJ/mol C 3 H 4
(b)
1000 g C 3H 4
1 m olC 3H 4
 1849.5 kJ


  4.616 104 kJ/kg C 3H 4
1 m olC 3H 4 40.065 g C 3H 4
1 kg C 3H 4
Propylene: C 3 H 6 (g) + 9/2 O 2 (g)  3 CO 2 (g) + 3 H 2 O(g)
(a)
(b)
  3(–393.5 kJ) + 3(–241.82 kJ) – (20.4 kJ) –9/2(0) = –1926.4
 H rxn
= –1926 kJ/mol C 3 H 6
1000 g C 3H 6
1 m olC 3H 6
 1926.4 kJ


  4.578 104 kJ/kg C 3H 6
1 m olC 3H 6 42.080 g C 3H 6
1 kg C 3H 6
Propane: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g)
(a)
(b)
(c)
5.96
  3(–393.5 kJ) + 4(–241.82 kJ) – (–103.8 kJ) –5(0) = –2044.0
 H rxn
= –2044 kJ/mol C 3 H 8
1000 g C 3H 8
1 m olC 3H 8
 2044.0 kJ


  4.635  104 kJ/kg C 3H 8
1 m olC 3H 8 44.096 g C 3H 8
1 kg C 3H 8
These three substances yield nearly identical quantities of heat per unit mass, but
propane is marginally higher than the other two.
   H  C O (g) 2 H  H O (g) H  C H (g) 2 H  O (g)
H rxn
f
2
f
2
f
4
f
2
= –393.5 kJ + 2(–241.82 kJ) – (–74.8 kJ) – 2(0) kJ = –802.3 kJ
   H  C F (g) 4  H  H F(g)  H  C H (g) 4  H  F (g)
 H rxn
f
4
f
f
4
f 2
= –679.9 kJ + 4(–268.61 kJ) – (–74.8 kJ) – 4(0) kJ = –1679.5 kJ
The second reaction is twice as exothermic as the first. The “fuel values” of hydrocarbons
in a fluorine atmosphere are approximately twice those in an oxygen atmosphere. Note
that the difference in H values for the two reactions is in the H º
f for the products,
because the H º
for
the
reactants
is
identical.
f
132
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5 Thermochemistry
5.97
Solutions to Exercises
Analyze/Plan. Given population, Cal/person/day and kJ/mol glucose, calculate kg
glucose/yr. Calculate kJ/yr, then kg/yr. 1 billion = 1  10 9 . 365 day = 1 yr. 1 Cal = 1
kcal, 4.184 kJ = 1 kcal = 1 Cal. Solve.
7.0  109 persons
1500 C al
365 day 4.184 kJ


 1.6035  1016  1.6  1016 kJ/yr
person-day
1yr
1C al
1.6035  1016 kJ 1m olC 6 12O 6 180.2 g C 6H 12O 6
1 kg



 1.0  1012 kg C 6 12O 6/yr
yr
2803 kJ
1 m olC 6 12O 6
1000 g
Check. 1  10 1 2 kg is 1 trillion kg of glucose.
5.98
(a)
Use density to change L to g, molar mass to change g to mol, heat of combustion
to change mol to kJ. Ethanol is C 2 H 5 OH , gasoline is C 8 H 1 8 . From Exercise 5.79
(c), heat of combustion of ethanol is –1234.8 kJ/mol.
1.0 L C 2H 5O H 
1000 m L 0.79 g 1m olC 2 5O H
1234.8 kJ



1L
1m L
46.07 g
1 m olC 2 5O H
= 21,174 = 2.1  10 4 kJ/L C 2 H 5 OH
1.0 L C 8H 18 
1000 m L 0.70 g
1m olC 8H 18
5400 kJ



1L
1 m L 114.23 g C 8H 18 1 m olC 8H 18
= 33,091 = 3.3  10 4 kJ/L C 8 H 1 8
(b)
If density and heat of combustion of E85 are weighted averages of the values for
the pure substances, than energy per liter E85 is also a weighted average of
energy per liter for the two substances.
kJ/L E85 = 0.15(kJ/L C 8 H 1 8 ) + 0.85(kJ/L C 2 H 5 OH)
kJ/L E85 = 0.15(33,091 kJ) + 0.85(21,174 kJ) = 22,962 = 2.3  10 4 kJ/L E85
(c)
Whether comparing gal or L, all conversion factors for the two fuels cancel, so we
can apply the energy ratio directly to the volume under consideration.
The energy ratio for E85 to gasoline is (22,962/33,091) = 0.6939 = 0.69
40 L gas 
(d)
kJ from E85
 57.65  58 L E85
0.6939 kJ from gas
If the E85/gasoline energy ratio is 0.69, the cost ratio must be 0.69 or less to
“break-even” on price. 0.69($3.88) = $2.68/gal E85
Check. 40 L  (1 gal/3.785 L)  gas($3.88/gal) = $41;
57.6 L  (1 gal/3.785 L)  E85($2.68/gal) = $41.
Additional Exercises
5.99
Like the combustion of H 2 (g) and O 2 (g) described in Section 5.4, the reaction that
inflates airbags is spontaneous after initiation. Spontaneous reactions are usually
exothermic, –H. The airbag reaction occurs at constant atmospheric pressure, H = q p ;
both are likely to be large and negative. When the bag inflates, work is done by the
system on the surroundings, so the sign of w is negative.
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5 Thermochemistry
Solutions to Exercises
5.100
Freezing is an exothermic process (the opposite of melting, which is clearly endothermic).
When the system, the soft drink, freezes, it releases energy to the surroundings, the can.
Some of this energy does the work of splitting the can. (When water freezes, it expands. It
is specifically this expansion that does the work of splitting the can.)
5.101
(a)
No work is done when the gas expands.
(b)
No work is done because the evacuated flask is truly empty. There is no
surrounding substance to be “pushed back.”
(c)
E = q + w. From part (b), no work is done when the gas expands. The flasks are
perfectly insulated, so no heat flows. E = 0 + 0 = 0. The answer is a bit
surprising, because a definite change occurred that required no work or heat
transfer and consequently involved no energy change.
(a)
q = 0, w > 0 (work done to system), E > 0
(b)
Because the system (the gas) is losing heat, the sign of q is negative.
5.102
Two interpretations of the final state in (b) are possible. If the final state in (b) is
identical to the final state in (a), E(a) = E(b). If the final volumes are identical,
case (b) requires either more (non-PV) work or heat input to compress the gas
because some heat is lost to the surroundings. (The moral of this story is that the
more energy lost by the system as heat, the greater the work on the system
required to accomplish the desired change.)
Alternatively, if w is identical in the two cases and q is negative for case (b), then
E(b) < E(a). Assuming identical final volumes, the final temperature and
pressure in (b) are slightly lower than those values in (a).
5.103
E = q + w = –11.73 kJ – 2.48 kJ = –14.21 kJ
H = q p = –11.73 kJ
5.104
Heat, in general, is not a state function and neither is heat at constant pressure. Only if
there is no other work being done than pressure-volume work, heat can be identified
with the change in enthalpy and only in this case, does it have a state function. Clearly,
the work involved in a fuel cell is not pressure-volume work, but electrical work, so
ΔH ≠ q, even though the two changes occur at constant pressure.
5.105
Find the heat capacity of 1.7  10 3 gal H 2 O.
C H 2O  1.7  103 galH 2O 
4 qt
1 103 cm 3
1g
1L
4.184 J




3
1 gal 1.057 qt
1L
1 g-°C
1cm
= 2.692  10 7 J/  C = 2.7  10 4 kJ/  C; then,
1kg
1 brick
2.692 107 J 1g-o C



 1.8  104 or 18,000 bricks
o
3
0.85 J 1 10 g 1.8 kg
1 C
Check. (1.7  ~16  10 6 )/(~1.6  10 3 )  17  10 3 bricks; the units are correct.
5.106
(a)
q Ag  0.233
J
 200 g Cu  (30.2  100.5)K   3, 276 J   3.276 KJ
g -K
The negative sign indicates the 3.276 kJ are lost by the Ag block.
134
Copyright © 2018 Pearson Education Ltd.
5 Thermochemistry
(b)
q water  4.184
Solutions to Exercises
J
 150 g Cu  (30.2  25.2)K  3,138 J  3.138 KJ
g -K
The positive sign indicates the 3.138 kJ are gained by the water.
(c)
The difference in the heat lost by the silver and the heat gained by the water is
–3,276 J + 3,138 J = –138 J. The temperature change of the calorimeter is 5°C. The
heat capacity of the (empty) calorimeter is
138 J
 27.6 J/K
5.0 K
(d)
Because q water is known to 1 decimal place, the difference has 1 decimal place and
the result has 1 sig fig.
If the rounded results from (a) and (b) are used,
J
q water  3, 276 J  4.184
 150 g  T
g -K
T  5.22 K = 5.22C
Tf  25.2 C + 5.22C  30.4C
5.107
(a)
From the mass of benzoic acid and its heat of combustion and rise in temperature
of the calorimeter, we cam calculate the heat capacity of the calorimeter.
q
C=

m
kJ
 0.47 g
g
 3.775 kJ/K
3.284 K
26.38
Heat produced when 0.53 g of caffeine is burnt:
3.775
kJ
 3.05 K  11.52 kJ
g
Heat produced when 1 mole of caffeine is burnt:
kJ
3.775  3.05 K 
K
(b)
g
mol  4, 219 kJ/mol
0.53 g
194.2
If the uncertainty in each temperature reading is 0.002 K, the uncertainty of each
temperature difference is about 0.003 K (by ‘squaric addition’ of absolute errors:
(0.0022 + 0.0022)½ = 0.0028). The relative error in measurement due to uncertainties
in mass and temperature measurement is then calculated by ‘squaric addition’ of
the following relative errors:
0.001 g 0.001 g 0.003 K 0.003 K
,
,
,
0.47 g 0.53 g 3.284 K 3.05 K
Relative error = (0.0022 + 0.0022 + 0.0012 + 0.0012)½ = 0.003 = 0.3 %, multiplied by
the result, 0.003  4,219 kJ/mol, yielding an error of 12.7 kJ/mol = 13 kJ/mol for the
molar heat of combustion of caffeine.
5.108
(a)
   H Fe(OH) (s)  H H (g)  H Fe(s)  2H H O(l)
 H rxn
f
2
f
2
f
f
2
kJ   kJ   kJ  
kJ 

   583.39
   0 mol    0 mol   2   285.83 mol   11.73 kJ/mol Fe
mol

 
 
 

(b)
Use the specific heat of water to calculate the energy required to heat the water.
135
Copyright © 2018 Pearson Education Ltd.
5 Thermochemistry
Solutions to Exercises
Use the density of water at 25°C to calculate the mass of water to be heated. Then,
use the ‘heat stoichiometry’ in (a) to calculate the mass of iron needed.
250 mL  0.997
8.343 kJ 
5.109
(a)
g
J
 4.184
 (30  22)K  8.343 kJ
mL
g-K
1 mol Fe 55.845 g Fe

 39.7 g Fe needed
11.73 kJ
1 mol Fe
For comparison, the equations have been balanced in such a way, that 1 mole of
acetylene is burned in each.
C2H2 (g) + 1/2 O2 (g)  2 C(s) + H2O(l)
C2H2 (g) + 3/2 O2 (g)  2 CO(g) + H2O(l)
C2H2 (g) + 5/2 O2 (g)  2 CO2 (g) + H2O(l)
(b)
  2H C(s)  H H O(l)  H C H (g)  1 H O (g)
H rxn
f
f
2
f
2 2
f
2
2
kJ  
kJ  1  kJ 
 kJ  
 2 0
   285.83
  226.77
 0


mol  
mol  2  mol 
 mol  
  512.6 kJ/mol C 2 H 2
  2H CO(s)  H H O(l)  H C H (g)  3 H O (g)
H rxn
f
f
2
f
2 2
f
2
2
kJ  
kJ  
kJ  3  kJ 

 2   110.5
  285.83
  226.77
 0


mol  
mol  
mol  2  mol 

  733.6 kJ/mol C2 H 2
  2H CO (g)  H H O(l)   H C H (g)  5 H O (g)
H rxn
f
2
f
2
f
2 2
f
2
2
kJ  
kJ  
kJ  5  kJ 

 2   393.5
  285.83
  226.77
 0


mol  
mol  
mol  2  mol 

  1, 299.6 kJ/mol C 2 H 2
(c)
5.110
Assuming that O2 (g) is present in excess, the reaction that produces carbon
dioxide represents the most negative ΔH per mole of acetylene burned. More of
the potential energy of the reactants is released as heat during the reaction to give
products of lower potential energy. The reaction that produces carbon dioxide is
the most “downhill” in enthalpy.
2[CH 4 (g)  2 O 2 (g)  CO 2 (g)  2 H 2 O(l)]
1
 [2 H 2 (g)  O 2 (g)  2 H 2 O(l)]
2
1
 [2 C2 H 6 (g)  7 O 2 (g)  4 CO 2 (g)  6 H 2 O(l)]
2
2 CH 4 (g)  C2 H 6 (g)  H 2 (g)
H  2( 890.3 kJ)  1, 780.6 kJ
1
H   [ 571.6 kJ]  285.8 kJ
2
1
H   [ 3120.8 kJ]
2
H   65.6 kJ
136
Copyright © 2018 Pearson Education Ltd.
5 Thermochemistry
5.111
Solutions to Exercises
For octane:
1 mol C8 H18
5.53 MJ
kJ

 48.4
1 mol C8 H18
114.23 g
g
1 mol C8 H18 0.70 g
5.53 MJ
kJ


 33.9
3
1 mol C8 H18
114.23 g
1 cm
cm3
For liquid butane:
1 mol C4 H10
2.88 MJ
kJ

 49.6
1 mol C4 H10
58.12 g
g
1 mol C4 H10 0.60 g
2.88 MJ
kJ


 29.8
1 mol C4 H10
58.12 g
1 cm3
cm3
For liquid hydrogen:
0.29 MJ 1 mol H 2
kJ

 144
1 mol H 2 2.016 g
g
0.29 MJ 1 mol H 2 0.07 g
kJ


 10.1
1 mol H 2 2.016 g 1 cm3
cm3
Liquid hydrogen provides the most energy per unit mass, liquid octane provides the
most energy per unit volume.
5.112
(a)
 Hrxn   Hf 1  hexene(l)   Hf cyclohexene(l)
kJ  
kJ 

   74
   156
 82 kJ/mol

mol  
mol 

(b)
Because the reaction is endothermic (H is positive), the product, 1‐hexene, has
more enthalpy than the reactant, cyclohexane.
(c)
Both Cyclohexane and 1‐hexene will burn to 6 moles of carbon dioxide and 6
moles of water, that is, both the reaction will end up at the same low enthalpy
level.
C 6 H 1 2 (g) + 9 O 2 (g)  6 CO 2 (g) + 6 H 2 O(l)
 Hf combustion products  6 Hf CO 2 (g)  6 Hf H 2 O(l)
kJ 
kJ 


 6 mol   393.5
 6 mol   285.83
  4, 076.0 kJ

mol 
mol 


The combustion of cyclohexane starts at a lower level (-156 kJ) than the
combustion of 1-hexene (–74 kJ), so this reaction is less exothermic. 1-hexene will
have the larger combustion enthalpy.
5.113
The reaction for which we want H is:
13
O2 (g)  4 CO2 (g) + 5 H2O(g)
2
Before we can calculate H for this reaction, we must calculate H f for C4 H10(l).
C4 H10 (l) +
We know that H f for C4 H10(g) is –124.7 kJ/mol, and that for C4 H10(l)  C4 H10(g),
H = 22.44 kJ/mol
Thus, H v ap = H f C4 H10(g) – H f C4 H10(l).
22.44 kJ/mol = –124.7 kJ/mol –H f C4 H10(l); H f C4 H10(l) = –147.1 kJ/mol
137
Copyright © 2018 Pearson Education Ltd.
5 Thermochemistry
Solutions to Exercises
Then for the overall reaction, the enthalpy change is:
  4  H CO (g)  5 H H O(l)  H C H (g)  13 H O (g)
 H rxn
f
2
f
2
f
4 10
f
2
2
kJ  
kJ  
kJ  13  kJ 

 4   393.5
 5  241.82
   147.1

0


mol  
mol  
mol  2  mol 

  2, 636 kJ/mol C2 H10
g
cm3
MJ
2, 636 kJ 1 mol C2 H10

 0.60 3  1000
  27
1 mol C 2 H10
58.12 g
L
L
cm
2 CH 3 OH(l) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(g)
kJ
    393.5 kJ   2   241.82 kJ     239 kJ   3  0 kJ    638
 H rxn








mol 
mol  
mol  2  mol 
mol CH3 OH


1 mol CH3OH
g
cm3
MJ
 638 kJ

 0.792 3 1000
  15.8
1 mol CH 3OH
32.04 g
L
L
cm
In terms of heat obtained per unit volume of fuel, liquid butane is significantly better
than methanol.
5.114
Butane, C 4 H 1 0 , MM = 58.12 g/mol
(a)
C 4 H 1 0 (g) + 13/2 O 2 (g)  4 CO 2 (g) + 5 H 2 O(l)
  4  H CO (g)  5  H H O(l)  H C H (g)  13 H O (g)
 H rxn
f
2
f
2
f
4 10
f
2
2
kJ  
kJ  
kJ  13  kJ 

 4   393.5
 5  285.83
  125

0


mol  
mol  
mol  2  mol 

kJ
  2,878
mol C4 H10
(b)
 2,878 kJ 1 mol C4 H10
kJ

 49.5
1 mol C4 H10
58.12 g
g
(c)
g 

10 1.008
mol 
%H 
 100%  17.3%
g
58.12
mol
1‐Butene, C 4 H 8 (g), MM = 56.11 g/mol
(a)
C 4 H 8 (g) + 6 O 2 (g)  4 CO 2 (g) + 4 H 2 O(l)
  4  H CO (g)  4 H H O(l)   H C H (g)  6  H O (g)
 H rxn
f
2
f
2
f
4 8
f
2
kJ 
kJ  
kJ   kJ 


 4   393.5
  4  285.83 mol     1 mol   6  0 mol 
mol



 
 

kJ
  2, 716
mol C4 H8
(b)
1 mol C4 H8
 2, 716 kJ

  48.4 kJ/g
1 mol C4 H8 56.11 g C4 H8
138
Copyright © 2018 Pearson Education Ltd.
5 Thermochemistry
(c)
Solutions to Exercises
g 

8  1.008
mol 
%H 
 100%  14.4%
g
56.11
mol
1‐Butane, C 4 H 6 (g), MM = 54.09 g/mol
(a)
C 4 H 6 (g) + 11/2 O 2 (g)  4 CO 2 (g) + 3 H 2 O(l)
  4 H CO (g)  3  H H O(l)  H C H (g)  11/2  H O (g)
 H rxn
f
2
f
2
f
4 6
f
2
kJ  
kJ  
kJ  7  kJ 

 4   393.5
  3  285.83 mol   165 mol   2  0 mol 
mol

 
 



kJ
  2,596
mol C4 H 6
(b)
1 mol C4 H 6
 2,596 kJ

  48.0 kJ/g
1 mol C4 H 6 54.09 g C4 H 6
(c)
g 

6  1.008
mol 
%H 
 100%  11.2%
g
54.09
mol
(d)
5.115
As the mass % H increases, the fuel value (kJ/g) of the hydrocarbon increases,
given the same number of C atoms.
E p = mgd. Be careful with units. 1 J 1 kg-m 2 /s2
201lb 
1kg
9.81m
45 ft 1yd
1m




 20 tim es
2.205 lb
tim e 3 ft 1.0936 yd
s2
 2.453  105 kg-m 2/s2  2.453  105 J 2.5  102 kJ
1 Cal = 1 kcal = 4.184 kJ
2.453  102 kJ
1C al
 58.63  59 C al
4.184 kJ
No, if all work is used to increase the man’s potential energy, 20 rounds of stairclimbing will not compensate for one extra order of 245 Cal fries. In fact, more than 59
Cal of work will be required to climb the stairs, because some energy is required to
move limbs and some energy will be lost as heat.
5.116
Plan. Use dimensional analysis to calculate the amount of solar energy supplied per m 2
in 1 h. Use stoichiometry to calculate the amount of plant energy used to produce
sucrose per m 2 in 1 h. Calculate the ratio of energy for sucrose to total solar energy, per
m 2 per h.
Solve. 1 W = 1 J/s, 1 kW = 1 kJ/s
1.0 kW
1.0 kJ/ s 1.0 kJ 60 s 60 m in 3.6× 103 kJ
=
= 2 ×
×
=
1h
m2
m2
m -s 1m in
m 2 -h
139
Copyright © 2018 Pearson Education Ltd.
5 Thermochemistry
Solutions to Exercises
5645 kJ
1m olsucrose 0.20 g sucrose
×
×
= 3.298 = 3.3 kJ/ m 2 -h forsucrose production
2
m olsucrose 342.3 g sucrose
m -h
3.298 kJ for sucrose
 100  0.092% sunlight used to produce sucrose
3.6  103 kJ totalsolar
5.117
(a)
6 CO 2 (g) + 6 H 2 O(l)  C 6 H 1 2O 6 (s) + 6 O 2 (g), H = 2803 kJ
This is the reverse of the combustion of glucose (Section 5.8 and Exercise 5.89), so
H = –(–2803) kJ = +2803 kJ.
5.5 1016 g C O 2 1 m olC O 2
2803 kJ


 5.838 1017  5.8 1017 kJ
yr
44.01 g C O 2 6 m olC O 2
(b)
1W  1 J/s;1W -s  1 J
5.838×1017 kJ 1000 J 1 yr 1d
1h
1m in 1 W -s
×
×
×
×
×
×
yr
kJ
365 d 24 h 60 m in 60 s
J
1M W

 1.851 107 M W  1.9 107 M W
1 106W
1.9 107 M W 
1 plant
103 M W
 1.9 104  19,000 nuclear power plants
Integrative Exercises
5.118
(a)
m
1, 609
mi
mile  431 m

965
s
h
s
3600
h
(b)
Find the mass of one N 2 molecule in kg.
g
kg
 103
mol
g
 6.64  1026 kg
1
23
6.022  10
mol
40
2
1
1
m

E = mv 2 =  6.64  1026 kg  431   6.17  1021 J
2
2
s 

5.119
1
 6.17  1021 J  3.17  103 J/mol = 3.17 kJ/mol
mol
(c)
6.022 1023
(a)
E p = mgd = 52.0 kg  9.81 m/s 2  10.8 m = 5509.3 J = 5.51 kJ
(b)
Ek  1/ 2 m v ; v  (2Ek /m )
(c)
Yes, the diver does work on entering (pushing back) the water in the pool.
(a)
CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l)
1/2
5.120
2
1/2
 2  5509.3 kg-m 2/s2 




52.0 kg


 14.6 m /s
 H o   H f C O 2 (g) 2  H f H 2O (l)  H f C H 4 (g) 2  H f O 2 (g)
= –393.5 kJ + 2(–285.83 kJ) – (–74.8 kJ) – 2(0) = –890.36 = –890.4 kJ/mol CH 4
The minus sign indicates that 890.4 kJ are produced per mole of CH 4 burned.
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5 Thermochemistry
Solutions to Exercises
890.36 kJ 1000 J
1 m ol


 1.4785  1018
23
m olC H 4
1 kJ 6.022  10 m olecules C H 4
 1.479  1018 J/m olecule
(b)
1eV = 96.485 kJ/mol
8 keV 
1000 eV 96.485 kJ
1 m ol
1000 J



 1.282 1015  1  1015 J/X -ray
1keV
eV -m ol 6.022  1023
kJ
The energy produced by the combustion of 1 molecule of CH 4 (g) is much smaller
than the energy of the X-ray.
5.121
(a,b) Ag + (aq) + Li(s)  Ag(s) + Li + (aq)
 H    H f Li (aq)  H f A g (aq)
= –278.5 kJ – 105.90 kJ = –384.4 kJ
Fe(s) + 2 Na + (aq)  Fe 2 +(aq) + 2 Na(s)
 H    H f Fe2 (aq) 2  H f N a (aq)
= –87.86 kJ – 2(–240.1 kJ) = +392.3 kJ
2 K(s) + 2 H 2 O(l)  2 KOH(aq) + H 2 (g)
 H   2  H f K O H (aq) 2  H f H 2O (l)
= 2(–482.4 kJ) – 2(–285.83 kJ) = –393.1 kJ
(c)
Exothermic reactions are more likely to be favored, so we expect the first and
third reactions be favored.
(d)
In the activity series of metals, Table 4.5, any metal can be oxidized by the cation
of a metal below it on the table.
Ag + is below Li, so the first reaction will occur.
Na + is above Fe, so the second reaction will not occur.
H + (formally in H 2 O) is below K, so the third reaction will occur.
These predictions agree with those in part (c).
5.122
(a)
 H   H f N aN O 3 (aq)  H f H 2O (l) H f H N O 3(aq) H f N aO H (aq)
H = –446.2 kJ – 285.83 kJ – (–206.6 kJ) – (–469.6 kJ) = –55.8 kJ
 H   H f N aC l(aq) H f H 2O (l) H f H C l(aq)  H f N aO H (aq)
H = –407.1 kJ – 285.83 kJ – (–167.2 kJ) – (–469.6 kJ) = –56.1 kJ
 H   H f N H 3(aq) H f N a (aq)  H f H 2O (l) H f N H 4 (aq)  H f N aO H (aq)
= –80.29 kJ – 240.1 kJ – 285.83 kJ – (–132.5 kJ) – (–469.6 kJ) = –4.1 kJ
(b)
H + (aq) + OH – (aq)  H 2 O(l)
(c)
The H values for the first two reactions are nearly identical, –55.8 kJ and
–56.1 kJ. The spectator ions by definition do not change during the course of a
reaction, so H is the enthalpy change for the net ionic equation. Because the
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5 Thermochemistry
5.123
Solutions to Exercises
(d)
first two reactions have the same net ionic equation, it is not surprising that they
have the same H.
Strong acids are more likely than weak acids to donate H + . The neutralizations of
the two strong acids are energetically favorable, whereas the neutralization of
NH 4 + (aq) is significantly less favorable. NH 4 + (aq) is probably a weak acid.
(a)
mol Cu = M  L = 1.00 M  0.0500 L = 0.0500 mol
g = mol  MM = 0.0500  63.546 = 3.1773 = 3.18 g Cu
(b)
The precipitate is copper(II) hydroxide, Cu(OH) 2 .
(c)
CuSO 4 (aq) + 2 KOH(aq)  Cu(OH) 2 (s) + K 2 SO 4 (aq), complete
Cu 2 +(aq) + 2 OH – (aq)  Cu(OH) 2 (s), net ionic
(d)
The temperature of the calorimeter rises, so the reaction is exothermic and the
sign of q is negative.
q   6.2 oC  100 g 
4.184 J
  2.6 103 J  2.6 kJ
1 g-o C
The reaction as carried out involves only 0.050 mol of CuSO 4 and the
stoichiometrically equivalent amount of KOH. On a molar basis,
H 
5.124
(a)
 2.6 kJ
  52 kJfor the reaction as written in part (c)
0.050 m ol
AgNO 3 (aq) + NaCl(aq)  NaNO 3 (aq) + AgCl(s)
net ionic equation: Ag + (aq) + Cl – (aq)  AgCl(s)
 H o   H f A gC l(s)  H f A g (aq)  H fC l (aq)
H = –127.0 kJ – (105.90 kJ) – (–167.2 kJ) = –65.7 kJ
(b)
H for the complete molecular equation will be the same as H for the net ionic
equation. Na + (aq) and NO 3 – (aq) are spectator ions; they appear on both sides of
the chemical equation. Because the overall enthalpy change is the enthalpy of the
products minus the enthalpy of the reactants, the contributions of the spectator
ions cancel.
(c)
 H o  H f N aN O 3(aq) H f A gC l(s) H f A gN O 3 (aq) H f N aC l(aq)
 H  A gN O (aq)  H  N aN O (aq)  H  A gC l(s) H  N aC l(aq)  H o
f
3
f
3
f
3
f
f
 H  A gN O (aq)  446.2 kJ ( 127.0 kJ) ( 407.1 kJ) ( 65.7 kJ)
 H f A gN O 3(aq)  100.4 kJ/m ol
5.125
(a)
21.83 g C O 2 
4.47 g H 2O 
12.01 g C
1 m olC O 2
1 m olC


 5.9572  5.957 g C
44.01 g C O 2 1 m olC O 2 1 m olC
1.008 g H
1 m olH 2O
2 m olH


 0.5001  0.500 g H
18.02 g H 2O 1 m olH 2O
m olH
The sample mass is (5.9572 + 0.5001) = 6.457 g
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5 Thermochemistry
(b)
Solutions to Exercises
5.957 g C 
1 m olC
 0.4960 m olC ; 0.4960/0.496  1
12.01 g C
0.500 g H 
1 m olH
 0.496 m olH ; 0.496/0.496  1
1.008 g H
The empirical formula of the hydrocarbon is CH.
(c)
Calculate "H of " for 6.457 g of the sample.
6.457 g sample + O 2 (g)  21.83 g CO 2 (g) + 4.47 g H 2 O(g),  H com b = –311 kJ
  H o C O (g)  H o H O (g) H o sam ple  H o O (g)
H com
b
f
2
f
2
f
f
2
o
o
o

H sam ple  H C O (g) H H O (g) H
f
f
2
f
H of C O 2 (g) 21.83 g C O 2 
H of H 2O (g) 4.47 g H 2O 
2
com b
1 m olC O 2  393.5 kJ

  195.185   195.2 kJ
44.01 g C O 2 m olC O 2
1 m olH 2O
 241.82 kJ

  59.985  60.0 kJ
18.02 g H 2O m olH 2O
 H of sam ple   195.185 kJ 59.985 kJ ( 311 kJ) 55.83  56 kJ
H of 
(d)
13.02 g
55.83 kJ

 112.6  1.1 102 kJ/C H unit
6.457 g sam ple C H unit
The hydrocarbons in Appendix C with empirical formula CH are C 2 H 2 and
C6H6.
substance
ΔH ºf / mol
ΔH ºf / CH unit
C 2 H 2 (g)
226.7 kJ
113.4 kJ
C 6 H 6 (g)
82.9 kJ
13.8 kJ
C 6 H 6 (l)
49.0 kJ
8.17 kJ
1.1  10 2 kJ
sample
The calculated value of H º
f /C H unit for the sample is a good match with
acetylene, C 2 H 2 (g).
5.126
(a)
CH 4 (g)  C(g) + 4 H(g)
(i) reaction given
CH 4 (g)  C(s) + 2 H 2 (g)
(ii) reverse of formation
The differences are: the state of C in the products; the chemical form, atoms, or
diatomic molecules, of H in the products.
(b)
i.
 H    H f C (g) 4  H f H (g)  H f C H 4 (g)
= 718.4 kJ + 4(217.94) kJ – (–74.8) kJ = 1665.0 kJ
ii.
H   H f C H 4  ( 74.8)kJ 74.8 kJ
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5 Thermochemistry
Solutions to Exercises
The rather large difference in H values is due to the enthalpy difference
between isolated gaseous C atoms and the orderly, bonded array of C atoms in
graphite, C(s), as well as the enthalpy difference between isolated H atoms and
H 2 molecules. In other words, it is due to the difference in the enthalpy stored in
chemical bonds in C(s) and H 2 (g) versus the corresponding isolated atoms.
(c)
CH 4 (g) + 4 F 2 (g)  CF 4 (g) + 4 HF(g)
H = –1679.5 kJ
The H value for this reaction was calculated in Solution 5.96.
3.45 g C H 4 
1.22 g F2 
1 m olC H 4
 0.21509  0.215 m olC H 4
16.04 g C H 4
1 m olF2
 0.03211  0.0321 m olF2
38.00 g F2
There are fewer mol F 2 than CH 4 , but 4 mol F 2 are required for every 1 mol of
CH 4 reacted, so clearly F 2 is the limiting reactant.
0.03211 m olF2 
5.127
 1679.5 kJ
  13.48   13.5 kJ heat evolved
4 m olF2
(a)
C 2 H 5 OH(l) + 3 O 2 (g)  2 CO 2 (g) + 3 H 2 O(l)
(b)
  2  H  C O (g)+ 3  H  H O (l) H  C H O H (l) 3  H  O (g)
H rxn
f
2
f
2
f
2 5
f
2
= 2(393.5) kJ + 3(285.83 kJ)  (277.7 kJ) 3(0) = 1366.79 = 1367 kJ
(c)
mL beer  mL ethanol  g ethanol
355 mL beer 
(d)
29.5735 mL 0.042 mL ethanol 0.789 g ethanol



oz
mL beer
mL
11.76 = 12 g ethanol
The metabolism reaction in part (a) produces 1367 kJ/mol ethanol, part (b).
11.76 g ethanol ×
(e)
1 mol ethanol
1367 kJ
1 kcal
×
×
= 83.40 = 83 kcal
46.07 g ethanol mol ethanol 4.184 kJ
Percent kcal from ethanol
83.40 kcal from ethanol
 100 = 75.82 = 76%
110 kcal total
144
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6
Electronic
Structure of Atoms
Visualizing Concepts
6.1
(a)
(b)
Analyze/Plan. We are given the speed of sound in dry air and the frequency of the
lowest audible wave. We must find the wavelength of the sound waves. For any
wave, speed equals wavelength times frequency.
Solve.
Speed =   ;
 = speed/20 Hz = 20 s—1 = 20/s
  sp eed /  ;
3 43 m
s

 1 7.1 5  1 7 m (to 1 sig fig, 20 m )
s
20
Analyze/Plan. We are asked to find the frequency of electromagnetic radiation with
this wavelength. All electromagnetic radiation in a vacuum has the speed 2.998  10 8
m/s; the symbol for the speed of light is “c”.  = c/.
Solve.
  c/  ;
   m
1

 1 .7  1 0 7 s 1  1 .7  1 0 7 H z
s
1 7.1 5 m
1 .7  1 0 7 H z 
(c)
6.2
6.3
6.4
1 MHz
  H z
 1 7 M H z (20 M H z , to 1 sig fig)
According to Figure 6.4, this would be radio frequency radiation.
Given: 2450 MHz radiation. Hz = s—1, unit of frequency. M = 1  106; 2450  106 Hz =
2.45  109 Hz = 2.45  109 s—1.
(a)
Find 2.45  109 s—1 on the frequency axis of Figure 6.4. The wavelength that
corresponds to this frequency is approximately 1  10—1 = 0.1 m or 10 cm.
(b)
No, visible radiation has wavelengths of 4  10—7 to 7  10—7 m, much shorter than 0.1 m.
(c)
Energy and wavelength are inversely proportional. Photons of the longer 0.1 m
radiation have less energy than visible photons.
(d)
Radiation of 0.1 m is in the low energy portion of the microwave region. The
appliance is probably a microwave oven. (Appliances with heating elements that glow
red or orange give off wavelengths in the visible or the near visible portion of the
infrared. The 0.1 m wavelength is too long to belong to these appliances.)
(a)
By inspection, wave (a) has the longer wavelength.
(b)
Wave (b) has the higher frequency because it has the shorter wavelength.
(c)
Wave (b) has a higher energy because it has a higher frequency (and shorter
wavelength).
(a)
(iii) Betelgeuse < (i) Sun < (ii) Rigel.
(b)
Black body radiation
146
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