Download Chapter 19 Homework Problems Answers

Chapter 19 Homework Problems
Answers
19.1
The term thermodynamics is meant to convey the two ideas of thermo (heat) and dynamics (motion),
namely movement or transfer of heat.
19.2
∆E = Efinal – Einitial
The first law of thermodynamics states that the internal energy may be transferred as heat or work, but it
cannot be created or destroyed. The change in internal energy of a system is the sum of two terms, the
amount of energy the system gains from heat transfer and the amount of energy the system gains from work
transfer: ∆E = q + w. ∆E is positive is energy flows into a system and negative if energy flows out of a
system.
19.3
If we refer to the equation in the answer to Review Question 19.2, we can see that if Efinal is larger than
Einitial, then ∆E must be positive, by definition. Thus, ∆E for an endothermic process is positive.
19.4
∆E is the change in internal energy, which is positive for an endothermic process. Heat, q, is the absorbed
by the system during the process, and it has a positive sign if the system absorbs heat during the process.
Work, w, is done on the system by the surroundings during the process, and it has a negative sign if the
system does work on the surroundings, whereas it has a positive sign if the surroundings do work on the
system during the process.
19.5
A state function is a thermodynamic quantity whose value is determined only by the state of the system
currently, and is not determined by a system's prior condition or history. A change in a state function is the
same regardless of the path that is used to arrive at the final state from the initial state. That is, changes in
state function quantities are path–independent. In the statement of the first law, only ∆E is a state function.
19.6
∆E is the heat of reaction at constant volume, and applies, for instance, to reactions performed in closed
vessels such as bomb calorimeters. ∆H is the heat of reaction at constant pressure.
19.7
A pressure in Pascals, by definition, has the units Newtons/m2, or N/m2.
Thus, P in pascals times ∆V in m3 is: N/m2 × m3 = N · m
The last result is the Newton · meter, or a force times a distance. By definition, a force of 1 N operating
over a distance of 1 meter is the quantity 1 joule. Thus, the quantity P × ∆V has the units of energy,
namely joules.
19.8
Since the quantity P∆V is negative (corresponding to a decrease in volume), ∆E must be a smaller negative
quantity than ∆H. ∆E is a smaller negative quantity than ∆H because, if the change is carried out at
constant pressure, some energy is gained by the system as volume contracts.
19.9
The most negative value of ∆E will have the most energy released.
(a)
Compressing the spring: energy is added to the system: w > 0
Heating the spring: energy is added to the system: q > 0
(b)
Expanding the spring: energy is removed from the system: w < 0
Cooling the spring: energy is removed from the system: q < 0
(c)
Compressing the spring: energy is added to the system: w > 0
Cooling the spring: energy is removed from the system: q < 0
(d)
Expanding the spring: energy is removed from the system: w < 0
Heating the spring: energy is added to the system: q > 0
In case (b), q < 0 and w < 0 so ∆E is the most negative
19.10
A spontaneous change, in thermodynamic terms, is one for which the sign of ∆G is negative. It is a process
that occurs by itself, without continued outside assistance. Kinetics plays no role in determining
spontaneity.
19.11
Student dependent answer. Examples of spontaneous changes may include: leaves falling from a tree, open
soda cans losing their carbonation, food molding in the refrigerator, dorm rooms becoming messy.
Examples of non–spontaneous changes may include: dorm rooms becoming neat and tidy, class notes
becoming organized, leaves being raked into a pile.
19.12
Student dependent answer. For the examples given above all of the non–spontaneous changes are
endothermic, they all require the surroundings expenditure of energy. The spontaneous changes are all
exothermic except, perhaps, the dorm room becoming messy. This change may be either exothermic or
endothermic.
19.13
A change that is characterized by a negative value for ∆H tends to be spontaneous. The only situation
where this is not true is that in which the value of the product T∆S is sufficiently negative to make the
quantity ∆G = ∆H – T∆S become positive.
19.14
Spontaneous processes tend to proceed from states of low probability to states of higher probability.
19.15
The ammonium and nitrate atoms are in a highly ordered geometry in the crystalline NH4NO3 sample.
When NH4NO3 dissolves, the NH4+ and NO3– ions become randomly dispersed throughout the solvent. The
increase in randomness that attends the dissolving of the solid is responsible for the process being
favorable, or spontaneous, in spite of the fact that the enthalpy change is endothermic.
19.16
Entropy is a measure of the number of microstates of a system. An equivalent statement is that entropy is a
measure of the statistical probability of a system.
19.17
Entropy is a measure of the randomness of a system, an increase in randomness corresponds to an increase
in entropy.
(a)
An increase in temperature allows the atoms and molecules to move more rapidly,
therefore there is an increase in entropy.
(b)
A decrease in volume limits the space that the molecules and atoms have to move in,
therefore there is a decrease in entropy.
(c)
Changing a liquid to a solid limits the motion of the molecules; therefore there is a
decrease in entropy.
(d)
Dissociating into individual atoms increases the entropy because there are more particles.
19.18
(a)
(d)
19.19
The statistical probability of a system in a given state, relative to all the other states, is the same regardless
of how the system happened to have been formed.
19.20
The entropy of the universe increases when a spontaneous event occurs.
19.21
A spontaneous event occurs when ∆G is negative. Since ∆G = ∆H – T∆S, even if the entropy is negative
for a process, the enthalpy factor, ∆H may be negative enough at a given temperature to allow the overall
change in free energy to be negative.
19.22
This is the statement that the entropy of a perfect crystalline solid at 0 K is equal to zero: S = 0 at 0 K.
19.23
The entropy of a mixture must be higher than that of two separate pure materials, because the mixture is
guaranteed to have a higher degree of disorder. Said another way, a mixture is more disordered than either
of its two separate components. Only a pure substance can have an entropy of zero, and then only at 0 K.
negative
negative
(b)
(e)
negative
negative
(c)
(f)
positive
positive
19.24
Entropy increases with increasing temperature because vibrations and movements within a solid lead to
greater disorder at the higher temperatures. Melting especially produces more disorder and vaporization
even more.
19.25
No, glass is an amorphous solid that is really a mixture of different substances so it is not a perfect
crystalline solid and does not have an entropy value of 0 at 0K.
19.26
∆G = ∆H – T∆S
19.27
(a)
(b)
(c)
A change is spontaneous at all temperatures only if ∆H is negative and ∆S is positive.
A change is spontaneous at low temperatures but not at high temperatures only if ∆H is negative
and ∆S is negative.
A change is spontaneous at high temperatures but not at low temperatures only if ∆H is positive
and ∆S is positive.
19.28
A change is nonspontaneous regardless of the temperature, if ∆H is positive and ∆S is negative.
19.29
The value of ∆G is equal to the maximum amount of work that may be obtained from any process.
19.30
A reversible process is one in which the driving force of the process is nearly completely balanced by an
opposing force. The situation is a bit esoteric, since no process can be run in a completely reversible
manner. Nevertheless, the closer one obtains to reversibility, the more efficient the system becomes as a
source of the maximum amount of useful work that can be achieved with the process.
19.31
The slower that the energy extraction is performed, the greater is the total amount of energy that can be
obtained. This is the same as saying that the most energy is available from a process that occurs reversibly.
19.32
As with other processes that are not carried out in a reversible fashion, the energy is lost to the environment
and becomes unavailable for use. In addition, much of the energy is lost as heat which is used to maintain
our body temperature.
19.33
A truly reversible process would take forever to occur. Thus, if we can observe an event happening, it
cannot be a truly reversible one.
19.34
At equilibrium, the value of ∆G is zero.
19.35
The process of bond breaking always has an associated value for ∆H that is positive. Also, since bond
breaking increases disorder, it always has an associated value for ∆S that is positive. A process such as
this, with a positive value for ∆H and a positive value for ∆S, becomes spontaneous at high temperatures.
19.36
Before the heat transfer, the molecules in the hot object vibrate and move more violently than do those of
the cooler object. When in contact with one another, the objects transfer heat through collisions, and
eventually, some of the kinetic energy of the molecules in the hot object is transferred to the molecules of
the cool object. This process of energy transfer continues until the objects have the same temperature. The
heat transfer is spontaneous because the scattering of kinetic energy among the molecules of both objects is
a process with a positive value for its associated ∆S.
19.37
See Figure 19.15.
19.38
Although a reaction may have a favorable ∆G, and therefore be a spontaneous reaction in the
thermodynamic sense of the word, the rate of reaction may be too slow at normal temperatures to be
observed.
19.39
As the temperature is raised, °G' will become less negative, if ∆H° is negative and ∆S° is negative.
Accordingly, less product will be present at equilibrium.
19.40
∆G = ∆G° + RT lnQ
19.41
∆G° = –RT lnK
18.42
The natural log of 1 is zero so ∆G° = 0.
19.43
The amount of energy needed to break all the chemical bonds in one mole of gaseous molecules to give
gaseous atoms.
19.44
It is easy to see why the conversion of a solid or liquid element to gaseous atoms is an endothermic process.
In the case of elements which exist naturally as gases, most are polyatomic (the exception are the noble
gases). To convert these elements to gaseous atoms will require an input of energy as bonds need to be
broken.
19.45
Heat of formation is defined as the amount of energy needed to form the compound from its elements in
their most stable state. C2(g) is not the naturally occurring state of carbon.
19.47
∆E = q + w
–1455 J = 812 J + w
w = –2267 J
Since w is defined to be the work done on the system by the surroundings, then in this case, a negative
amount of work is done on the system by the surroundings. The system, in fact, does work on the
surroundings.
19.53
13
O2(g) 4CO2(g) + 5H2O(g)
2
Find the moles of C4H10 at the first set of pressure, volume, and temperature data, then find the moles of the
CO2 and H2O resulting from the reaction. Finally, find the new volume of the products.
PV = nRT
( 2.00 atm )(10.0 L )
mol C4H10 =
= 0.826 mol C4H10
( 0.0821 L atm mol−1 K −1 ) ( 295 K )
C4H10(g) +
 4 mol CO 2 
mol CO2 = ( 0.826 mol C 4 H10 ) 
 = 3.30 mol CO2
 1 mol C4 H10 
 5 mol H 2 O 
mol H2O = ( 0.826 mol C 4 H10 ) 
 = 4.13 mol H2O
 1 mol C4 H10 
total moles of product = 3.30 mol CO2 + 4.13 mol H2O = 7.43 mol
( 0.826 mol ) ( 0.0821 L atm mol−1 K −1 ) ( 301 K )
initial volume =
= 20.4 L
1.00 atm
final volume =
19.57
19.59
1.00 atm
= 184 L
w = –P∆V
w = –(1.00 atm)(184 L – 20.4 L) = 164 L atm
 101.325 J  1 kJ 
In kilojoules: 164 L atm 

 = 16.6 kJ
 1 L atm  1000 J 
2HI(g) H2(g) + I2(g)
The factors needed to be consider in order to determine the sign of ∆S are
(1)
the number of moles of products versus reactants
in this case the number of moles are the same
(2)
the state of the products versus reactants
both products and reactants are gases
(3)
the complexity of the molecules
HI is more complex than either H2 or I2
∆S is expected to be negative.
(a)
(b)
(c)
(d)
19.61
( 7.43 mol ) ( 0.0821 L atm mol−1 K −1 ) ( 301 K )
∆S is positive since randomness in a gas is higher than that in a solid.
∆S is negative. There are fewer moles of gases among the products.
∆S is negative since gaseous material (which is highly random) is replaced by a solid (which is
highly ordered).
∆S is negative since the relatively random liquid reactant disappears in a process that makes only a
solid.
∆S° = (sum S°[products]) – (sum S°[reactants])
(a)
∆S° = {S°[AgCl(s)]} – {1/2S°[Cl2(g)] + S°[Ag(s)]}
∆S° = {1 mol × (96.2 J mol–1 K–1)}
– {1/2 mol × (223.0 J mol–1 K–1) + 1 mol × (42.55 J mol–1 K–1)}
∆S° = –57.9 J/K
(b)
∆S° = {S°[H2O(g)]} – {1/2S°[O2(g)] + S°[H2(g)]}
∆S° = {1 mol × (188.7 J mol–1 K–1)}
– {1/2 mol × (205.0 J mol–1 K–1) + 1 mol × (130.6 J mol–1 K–1)}
∆S° = –44.4 J/K
(c)
∆S° = {S°[H2O(l)]} – {1/2S°[O2(g)] + S°[H2(g)]}
∆S° = {1 mol × (69.96 J mol–1 K–1)}
– {1/2 mol × (205.0 J mol–1 K–1) + 1 mol × (130.6 J mol–1 K–1)}
∆S° = –163.1 J/K
(d)
∆S° = {S°[CO2(g)] + S°[H2O(g)] + S°[CaSO4(s)]}
– {S°[CaCO3(s)] + S°[H2SO4(l)]}
∆S° = {1 mol × (213.6 J mol–1 K–1) + 1 mol × (188.7 J mol–1 K–1)
+ 1 mol × (107 J mol–1 K–1)} – {1 mol × (92.9 J mol–1 K–1)
+ 1 mol × (157 J mol–1 K–1)}
∆S° = +259 J/K
∆S° = {S°[NH4Cl(s)]} – {S°[HCl(g)] + S°[NH3(g)]}
∆S° = {1 mol × (94.6 J mol–1 K–1)}
– {1 mol × (186.7 J mol–1 K–1) + 1 mol × (192.5 J mol–1 K–1)}
∆S° = –284.6 J/K
∆S° = (sum S°[products]) – (sum S°[reactants])
∆S° = {S°[HC2H3O2(l)] + S°[H2O(l)]} – {S°[C2H5OH(l)] + S°[O2(g)]}
∆S° = {1 mol × (160 J mol–1 K–1) + 1 mol × (69.96 J mol–1 K–1)}
– {1 mol × (161 J mol–1 K–1) + 1 mol × (205.0 J mol–1 K–1)}
∆S° = –136 J/K
(e)
19.65
19.66
The quantity ∆G of applies to the equation in which one mole of pure phosgene is produced from the
naturally occurring forms of the elements:
C(s) + 1/2O2(g) + Cl2(g) COCl2(g), ∆G of = ?
We can determine ∆G of if we can find values for ∆Hof and ∆Sof , because:
∆G° = ∆H° – T∆S°
The value of ∆Sof is determined using S° for phosgene in the following way:
∆Sof = {S°[COCl2(g)]} – {S°[C(s)] + 1/2S°[O2(g)] + S°[Cl2(g)]}
∆Sof = {1 mol × (284 J mol–1 K–1)} – {1 mol × (5.69 J mol–1 K–1)
+ 1/2 mol × (205.0 J mol–1 K–1) + 1 mol × (223.0 J mol–1 K–1)}
∆Sof = –47 J mol–1 K–1 or –47 J/K
∆G of = ∆Hof – T ∆Sof = –223 kJ/mol – (298 K)(–0.047 kJ/mol K)
= –209 kJ/mol
19.71
Add the reverse of the first equation to the second equation plus twice the third equation:
CO(NH2)2(s) + 2NH4Cl(s) COCl2(g) + 4NH3(g),
COCl2(g) + H2O(l) CO2(g) + 2HCl(g),
2NH3(g) + 2HCl(g) 2NH4Cl(s),
∆G° = +332.0 kJ
∆G° = –141.8 kJ
∆G° = –183.9 kJ
CO(NH2)2(s) + H2O(l) 2NH3(g) + CO2(g),
∆G° = +6.3 kJ
19.73
We must first determine ∆G° for the reaction:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
∆G° = (sum ∆G of [products]) – (sum ∆G of [reactants])
∆G° = { ∆G of [CO2(g)] + 2 ∆G of [H2O(g)]} – { ∆G of [CH4(g)] + 2 ∆G of [O2(g)]}
∆G° = {1 mol × (–394.4 kJ/mol) + 2 mol × (–228.6 kJ/mol)}
– {1 mol × (–50.79 kJ/mol) + 2 mol × (0.0 kJ/mol)}
∆G° = –800.8 kJ/mol
19.76
Next, we determine the amount of work available from the combustion of 48.0 g of CH4:
 1 mol CH 4  −800.8 kJ 
3
kJ = ( 48.0 g CH 4 ) 

 = − 2.40 × 10 kJ
 16.043 g CH 4  1 mol CH 4 
At equilibrium, ∆G = 0 = ∆H – T∆S
Thus ∆H = T∆S, and if we assume that both ∆H and ∆S are independent of temperature, we have:
∆S = ∆H/Teq = (37.7 × 103 J/mol) ÷ (99.3 + 273.15 K)
∆S = 101 J mol–1 K–1
19.77
We proceed as in the answer to Review Problem 19.76:
∆S = ∆H/Teq = (31.9 × 103 J/mol) ÷ (56.2 + 273.15 K)
∆S = 96.9 J mol–1 K–1
19.79
We first balance each equation, and then calculate a value of ∆G°. If ∆G° is a negative number, then the
reaction is spontaneous.
(a)
3PbO(s) + 2NH3(g) 3Pb(s) + N2(g) + 3H2O(g)
∆G° = {3 ∆G of [Pb(s)] + ∆G of [N2(g)] + 3 ∆G of [H2O(g)]}
– {3 ∆G of [PbO(s)] + 2 ∆G of [NH3(g)]}
∆G° = {3 mol × (0.0 kJ/mol) + 1 mol × (0.0 kJ/mol)
+ 3 mol × (–228.6 kJ/mol)} – {3 mol × (–189.3 kJ/mol)
+ 2 mol × (–16.7 kJ/mol)}
∆G° = –84.5 kJ ∴ the reaction is spontaneous.
(b)
NaOH(s) + HCl(g) NaCl(s) + H2O(l)
∆G° = { ∆G of [NaCl(s)] + ∆G of [H2O(l)]} – { ∆G of [NaOH(s)]+ ∆G of [HCl(g)]}
∆G° = {1 mol × (–384.0 kJ/mol) + 1 mol × (–237.2 kJ/mol)}
– {1 mol × (–382 kJ/mol) + 1 mol × (–95.27)}
∆G° = –144 kJ ∴ the reaction is spontaneous.
(c)
Al2O3(s) + 2Fe(s) Fe2O3(s) + 2Al(s)
∆G° = { ∆G of [Fe2O3(s)] + 2 ∆G of [Al(s)]} – { ∆G of [Al2O3(s)] + 2 ∆G of [Fe(s)]}
∆G° = {1 mol × (–741.0 kJ/mol) + 2 mol × (0.0 kJ/mol)}
– {1 mol × (–1576.4 kJ/mol) + 2 mol × (0.0 kJ/mol)}
∆G° = +835.4 k ∴ the reaction is not spontaneous.
2CH4(g) C2H6(g) + H2(g)
∆G° = { ∆G of [C2H6(g)] + ∆G of [H2(g)]} – {2 ∆G of [CH4(g)]}
∆G° = {1 mol × (–32.9 kJ/mol) + 1 mol × (0.0 kJ/mol)}
– {2 mol × (–50.79 kJ/mol)}
∆G° = +68.7 kJ ∴ the reaction is not spontaneous.
(d)
19.83
79.8 × 103 J = –(8.314 J/K mol)(673 K) × ln Kp
ln Kp = –14.3 ∴ Kp = 6.40 × 10–7
We start by calculating the reaction quotient, Q. Be sure to determine the pressure of the gases using the
ideal gas law.
 3.8 × 10−3 mol 0.0821 L atm ( 673 K ) 
mol K




2.50 L


Q=
= 0.195
 ( 0.040 mol ) 0.0821 L atm ( 673 K )   ( 0.022 mol ) 0.0821 L atm ( 673 K ) 
mol K
mol K



2.50 L
2.50 L






(
)(
(
)
)
(
)
Since the value of Q is larger than the value of Kp, the system must shift to the left in order to reach
equilibrium.
19.91
∆Hof [C2H4(g)] refers to the enthalpy change under standard conditions for the following reaction:
2Cgraphite + 2H2(g) C2H4(g),
∆Hof [C2H4(g)] = 52.284 kJ/mol
We can arrive at this net reaction in an equivalent way, namely, by vaporizing all of the necessary elements
to give gaseous atoms, and then allowing the gaseous atoms to form all of the appropriate bonds. The
overall enthalpy of formation by this route is numerically equal to that for the above reaction, and,
conveniently, the enthalpy changes for each step are available in either Table 19.3 or Table 19.4:
∆Hof = sum( ∆Hof [gaseous atoms]) – sum(average bond energies in the molecule)
∆Hof [C2H4(g)] = 52.284 kJ/mol = [2 × 716.7 + 4 × 218.0] – [4 × 412 + C=C] from which we can calculate
the C=C bond energy: 605 kJ/mol.
19.105 We can calculate the work of the expanding gas (P∆V) if we can calculate the change in volume ∆V. Since
the initial volume is given (5.00 L), we need only to calculate the final volume. For this, it is first
necessary to determine the value of n, the number of moles of gas.
n=
PV
(4.00 atm)(5.00 L)
=
= 0.817 mol
L atm (298 K)
RT
0.0821 mol
K
V2 =
(
)
(
(0.817 mol) 0.0821
nRT
=
P2
1 atm
L atm
mol K
) (298 K)
= 20.0 L
∆V = (20.0 L − 5.00 L) = 15.0 L
The work of gas expansion against a constant pressure of 1 atm is then given by the quantity –P∆V: w = –
(1 atm)(15.0 L) = –15.0 L atm
19.119 We need to calculate the amount of energy produced when one gallon of each of these fuels is burned;
 0.7893 g ethanol   1 mole ethanol 
mol ethanol = 3.78 × 103 mL ethanol 
 = 64.8 moles ethanol

 1 mL ethanol   46.07 g ethanol 
(
)
 −1299.8 kJ 
4
kJ = (64.8 moles ethanol) 
 = 8.42 × 10 kJ
 1 mole ethanol 
 0.7025 g octane   1 mole octane 
mol octane = 3.78 × 103 mL octane 
 = 23.2 moles octane

 1 mL octane   114.23 g octane 
(
)
 −5307 kJ 
5
kJ = (23.2 moles octane) 
 = 1.23 × 10 kJ
1
mole
octane


In spite of the large number of moles of ethanol in one gallon of liquid, the energy produced from the
combustion of a gallon of octane is greater than the amount produced when one gallon of ethanol is burned.