CHAPTER 2
1. a) q p = ΔH = 1 L × 0.9982 kg L–1 × 2447 kJ kg–1 = 2443 kJ
b) ΔT = 2443 kJ/(60 kg × 4.184 kJ K –1 kg–1 ) = 9.7 K
c) C12 H 22 O 11 (s) + 12O 2 (g) → 12 CO 2 (g) + 11H 2 O (l)
ΔH = 11(–285.83 kJ mol–1 ) + 12(–393.51 kJ mol–1 ) – (–2222.10 kJ mol–1 )
= –5644 kJ mol–1
Msucrose = 342.31 g mol–1
msucrose = 2443 kJ × 342.31 g mol–1 /5644 kJ mol–1 = 148.2 g
2. a) 1 kg of carbon is (1.0 kg × 1000 g kg–1 )/(12.011 g mol–1 ) = 83.3 mol
83.3 mol CO 2 × 8.3145 J mol–1 K –1 × 298 K/(105 Pa) = 2.06 m3 CO 2
2.06 m3 CO 2 /(0.000390 m3 CO 2 / m3 air ) = 5292 m3 air
b) p = F/A = mg/A = 105 Pa. m = Ap/g = 1 m2 × 105 Pa/(9.80662 m s –2 ) = 10197 kg
Mair = 0.8 × 0.0280 kg mol–1 + 0.2 × 0.0320 kg mol–1 = 0.0288 kg/(mol air)
nair = 10197 kg/0.0288 kg/(mol air) = 354069 mol air
nCO2 = 354069 mol air × 0.000390 mol CO 2 /(mol air) = 138.09 mol CO 2
mC = 0.012011 kg/mol C × 138.09 mol CO 2 = 1.659 kg C
c) 1.659 kg C/(1 kg y –1 ) = 1.659 y.
3. a) w = mgh = 10 kg × 9.81 m s –2 × 10 m = 981 J
b) k = F/(x – x0 ) = 5.00 N/(0.105 m – 0.100 m) = 1.00 × 103 N m–1
w = (k/2)(x – x0 ) 2 = 1.00 × 103 N m–1 × (0.01 m) 2 /2 = 0.050 J
Δd/dsurface × 100% = 1.2%
c) w = –p(V2 – V1 ) = –105 Pa × 0.002 m3 = –200 J.
d) w = –2 × 10–4 J
e) w

CHAPTER 2
1. a) q p = ΔH = 1 L × 0.9982 kg L–1 × 2447 kJ kg–1 = 2443 kJ
b) ΔT = 2443 kJ/(60 kg × 4.184 kJ K –1 kg–1 ) = 9.7 K
c) C12 H 22 O 11 (s) + 12O 2 (g) → 12 CO 2 (g) + 11H 2 O (l)
ΔH = 11(–285.83 kJ mol–1 ) + 12(–393.51 kJ mol–1 ) – (–2222.10 kJ mol–1 )
= –5644 kJ mol–1
Msucrose = 342.31 g mol–1
msucrose = 2443 kJ × 342.31 g mol–1 /5644 kJ mol–1 = 148.2 g
2. a) 1 kg of carbon is (1.0 kg × 1000 g kg–1 )/(12.011 g mol–1 ) = 83.3 mol
83.3 mol CO 2 × 8.3145 J mol–1 K –1 × 298 K/(105 Pa) = 2.06 m3 CO 2
2.06 m3 CO 2 /(0.000390 m3 CO 2 / m3 air ) = 5292 m3 air
b) p = F/A = mg/A = 105 Pa. m = Ap/g = 1 m2 × 105 Pa/(9.80662 m s –2 ) = 10197 kg
Mair = 0.8 × 0.0280 kg mol–1 + 0.2 × 0.0320 kg mol–1 = 0.0288 kg/(mol air)
nair = 10197 kg/0.0288 kg/(mol air) = 354069 mol air
nCO2 = 354069 mol air × 0.000390 mol CO 2 /(mol air) = 138.09 mol CO 2
mC = 0.012011 kg/mol C × 138.09 mol CO 2 = 1.659 kg C
c) 1.659 kg C/(1 kg y –1 ) = 1.659 y.
3. a) w = mgh = 10 kg × 9.81 m s –2 × 10 m = 981 J
b) k = F/(x – x0 ) = 5.00 N/(0.105 m – 0.100 m) = 1.00 × 103 N m–1
w = (k/2)(x – x0 ) 2 = 1.00 × 103 N m–1 × (0.01 m) 2 /2 = 0.050 J
Δd/dsurface × 100% = 1.2%
c) w = –p(V2 – V1 ) = –105 Pa × 0.002 m3 = –200 J.
d) w = –2 × 10–4 J
e) w = –nRTln(V2 /V1 ) = –p1V1 ln(V2 /V1 ) = –105 Pa × 10–3 m3 × ln 3 = –110 J.
4. a) w = f dx =
0
10
∫ (20x + x2 ) dx =
0
10
∫ 10x2 + 1
3 x3
0
10
= (1000 + 333)

Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry
Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak
2
= 0.0375 L. Density is reciprocally related to volume. 10/0.0375 = 266.9 – 1 = 265.9 =
26590%
6. nH 2 O = 100 g/(18.015 g mol–1 ) = 5.55 mol.
a) q = 5.55 mol × 75.4 J mol–1 K –1 × 100 K = 4.19 × 104 J
b) q = nH 2 OΔfusH = 5.55 mol × 6007 J mol–1 = –3.33 × 104 J
c) q = nH 2 OΔvapH = 5.55 mol × 40667 J mol–1 = 2.26 × 105 J
7. a) V1 = 1 mol × 8.3145 J mol–1 K –1 × 300 K /(105 Pa) = 0.0249 m3
V2 = 1 mol × 8.3145 J mol–1 K –1 × 600 K /(10 5 Pa) = 0.0499 m3
wp = –p(V2 – V1 ) = –10 5 Pa × (0.0499 – 0.0249)m3 = –2494 J
b) ΔU = nCV,m ΔT = 1 mol × 20.8 J mol–1 K –1× 300 K = 6.24 kJ
ΔH = ΔU + Δ(pV) = 6.24 kJ + 2.49 kJ = 8.73 kJ
c) q p = ΔH = 8.73 kJ
8. a) The volume, V = (4/3)πr3 = 1.333 × 3.142 × (5 × 10–7 m) 3 = 5.236 × 10–19 m3
m = Vρ = 5.236 × 10–19 m3 × 1000 kg m–3 = 5.236 × 10–16 kg.
b) M = mNA = 5.236 × 10–16 kg × 6.022 × 1023 mol–1 = 3.153 × 108 kg mol–1 (kD)
c) Nwater = M/Mwater = 3.153 × 108 kg mol–1/0.018 kg mol–1 = 1.752 × 1010
d) 6378.1 × 103 m/(1 × 10 –6 m) = 6.3781 × 1012
e) A = 4πr2 = 4 × 3.142 × (5 × 10–7 m) 2 = 3.142 × 10–12 m2
f) 3.142

Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry
Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak
3
w < 0, expansion
q = –w > 0
c) q = 0, adiabatic.
w = 0, expansion against zero pressure
ΔU = w + q = 0.
ΔH = ΔU + RΔT = 0
d) q > 0, heat of vaporization.
w < 0, expansion against pressure
ΔH = q p > 0.
ΔU = ΔH – Δ(pV) > 0; ΔH >>Δ(pV) ≅ ngRT
e) q < 0, exothermic reaction.
w = 0, closed bomb, ΔV = 0
ΔU = q V < 0.
ΔH = q V + Δ(pV) < 0; Δ(pV) = –(nH 2 + nO 2 )RT < 0
11. a) ΔH = q p = 40.667 kJ.
ΔU = ΔH – Δ(pV ) = ΔH – Δ(ngRT ) = 40667 J – (1 mol × 8.3145 J mol–1 K –1 × 373.15 K)
= 37.564 kJ
wp = pΔV = Δ(ngRT ) = 1 mol × 8.3145 J mol–1 K –1 × 373.15 K = –3.103 kJ
b) w = w1 + w2 = 0, since pex = 0 in step 1 and ΔV = 0 in step 2.
q = ΔU = 37.564 kJ, path independent.
ΔH = 40.667 kJ, path independent.
12. a) q = nCm (T2 − T1 ) = (1 mol)(38.0 J K−1 mol−1 )(20 K) = 760 J
b) q = (1 mol)(6007 J mol−1 ) = 6007 J
c) Per degree temperature drop, 10 mol H 2 O (l) loses
(10 mol)(75.4 J mol−1
K−1 ) = 754 J K−1
760 J + 6007 J + (1 mol)(75.4 J K−1 mol−1 )(T f − 0) = (10 mol)(75.4 J K−1
mol−1 )(20 − T f )
T

Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry
Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak
4
b) ΔT = − 400 J (1 mol × 75.95 J mol−1 K−1 ) = −5.27 °C
T f = 100 − 5.27 = 94.73°C
V f (l) = (18 g)/(0.96 g mL−1 ) = 18.75 mL . No phase change; ∆V is negligible for the
liquid water.
c) Δng = −400 J( ) 40,657 J mol−1
( ) = −9.84 × 10−3 mol
∆ng mol water vapor will condense to liquid
No temperature change
ΔV = ΔnRT / p = (−9.84 × 10−3 mol)(8.314 J mol-1 K -1 )(373 K) / (105 Pa)
ΔV = −3.05 × 10−4 m 3
d) Part (a), because it has the greatest ΔV; w > 0.
14. V = 5.8 – 2 × exp[100(0.002)] =3.36 L
At 1.005 bar, V = 5.8 – 2 × exp[100(–0.005)] =4.59 L
Δn = 105 Pa × (4.59 – 3.36) × 10–3 m3 /(8.3145 J mol–1 K –1 × 310 K) = 0.0477 mol
15. a) q = 0, thermally insulated
w = 0, assuming no volume change because decrease in volume of hot copper
equals increase in cool copper.
ΔU = 0, energy is conserved.
ΔH = 0
b) q = 0, thermally insulated
w = 0, mechanical work done on the system
ΔU = w > 0
ΔH = 0 because ΔpV = 0, with negligible expansion of the system.
c) q = 0
w = 0
ΔU = 0
ΔH = 0 (all answers assume ideal gas behavior)
16. a) First Law of Thermodynamics – no restrictions
b) Constant pressure process with no non-pV work.
c) Enthalpy is a linear function of temperature (heat capacity is constant). i.e. ΔH/ΔT =
dH/dT.
d) Reactions in ideal gases at constant temperature, where Δ(pV) = RTΔn
e) van der Waals gas.
f) Constant pex .

Loading page 6...

Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry
Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak
6
Δr H ° = (−285.830) + 0 − (−191.17)⎡⎣ ⎤⎦ kJ mol−1 = −94.66 kJ mol−1
A temperature rise of 0.02 ˚C requires (0.02 ˚C)(4.184 kJ kg–1)(1 kg L–1 ) = 0.0837 kJ L–1
Minimum detectableconcentration = 0.0837 kJ L−1
94.66kJ mol−1 = 8.84 × 10−4 M
24. a) ΔH °298 = −74.81( ) + 0 − −

Loading page 7...

Loading page 8...

Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry
Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak
8
31. To measure Δr H298
° the reactants could be put into a calorimeter at 25 ˚C and a small amount of
an enzyme catalyst added. The heat evolved per mole at constant pressure is Δr H298
° To estimate
ΔrU298
° the volume change of the reaction must be determined. The change in volume can be
measured in a dilatometer, or partial molal volumes can be used to calculate the change in
volume. For practical purposes the volume change is negligible, since no gases are involved, and
ΔrU298
° = Δr H298
°
.
32. a) 22 kWhd−1
( ) 3600 kJ (kWh)−1
( ) = 7.92 × 10 4 kJ d−1
b) 1 kW m−2
( ) 0.1( ) 5 h( ) = 0.5 kWh m−2
A = 22 kWh( ) 0.5 kWh m−2
( ) = 44 m 2
33. a) Δr H298
° = 2Δf H ° H 2 O, g( ) + Δf H ° O 2 , g( ) − Δf H ° H 2 O 2 , g( )
= 2 −241.818( ) + 0 − 2 −133.18( )⎡⎣ ⎤⎦ kJ mol−1 = −217.28 kJ mol−1
b)
( ) ( ) ( ) 1
molkJ5.1403.4982128.21721OOD −
=+−=−
c) Δr H °298 = 2 −285.830( ) + 0 − 2 −191.17( )⎡⎣ ⎤⎦ kJ mol−1 = −189.32 kJ mol−1
d) Heat evolved = 1 2 189.32 kJ (mol H 2 O 2 )−1
( ) 0.01 molkg−1
( ) = 0.947 kJ kg−1
Temperature rise = 0.947 kJ kg−1
( ) 4.18 kJ K

Loading page 9...

Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry
Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak
9
Chief discrepancy lies in using average bond dissociation energies for D(C—C) and
D(C—H)
b) 10 C (graphite) + 4 H 2 (g) →
Δf H ° = 10Δf H ° C( ) + 4D H − H( ) − 6D C − C( ) − 5D C = C( ) − 8D C − H( )
= 10 716.7( ) + 4 436.0( ) − 6 359( ) − 5 611( ) − 8 411( )⎡⎣ ⎤⎦ kJ mol−1
= 414 kJ mol−1
Published value 1
molkJ96.150 −
Chief discrepancy arises from neglect of resonance energy.
c) C graphite( ) + H 2 g( ) + 1
2 O 2 g( ) → H 2 C = O g( )
Δf H ° = Δf H ° C( ) + D H − H( ) + 1
2 D O 2( ) − 2D C − H( ) − D C = O( )
= 716.7 + 436.0 + 498.3 / 2 − 2 411( ) − 709 = −129 kJ mol−1
Published value 1
molkJ90.115 −
−
Chief discrepancy arises from use of average bond dissociation energies.
d) C (graphite) + H 2 (g) + O 2 (g) → formic acid
ΔH f
° = ΔH f C( ) + D H − H( ) + D O 2( ) − D C − H( ) − D C = O( ) − D C − O( ) − D O − H( )
= 716.7 + 436.0 + 498.3

Loading page 10...

3-14
CHAPTER 3
1. a) path 1 : (1 mole ideal gas, 3 bar, 450K)  (1 mol, 3 bar, 250 K) (1 mol, 2 bar, 250 K)
path 2 : (1 mole ideal gas, 3 bar, 450K)  (1 mol, 2 bar,450 K) (1 mol, 2 bar, 250 K)
path 1: the first step is isobaric, the second is isothermal; both reversible
path 2: the first step is isothermal, the second is isobaric; both reversible
work(path 1) =- nR(250-450K)+ nR(250K)ln(2bar 3bar)
= -(1mol×8.314J mol−1 K−1)(−200K)
+ (1mol×8.314J mol−1K−1)(250K)ln(2bar 3bar)
= 820.0408 J
q(path 1) = nC p(250− 450K)− nR(250K)ln(2bar 3bar)
= (1mol× 2.5*8.314J mol−1 K−1)(−200K)
− (1mol×8.314J mol−1K−1)(250)ln(2 3)
= −3314.2408 J
work(path 2) = nR(450K)ln(2bar 3bar)- nR(250-450K)
= (1mol×8.314J mol−1 K−1)(450K)ln(2bar 3bar)
− (1mol×8.314J mol−1K−1)(−200)
=145.8334 J
q(path 2) = −nR(450K)ln(2bar 3bar)+ nC p(250− 450K)
= −(1mol×8.314J mol−1 K−1)(450K)ln(2bar 3bar)
+ (1mol× 2.5×8.314J mol−1 K−1)(−200K)
= −2640.0334 J
(excess digits are shown since values will be rounded in part b)
b) ΔU (path 1) = qpath 1 + wpath 1 = 820.0408− 3314.2408 = −2494.2 J (first law)
ΔU (path 2) = qpath 2 + wpath 2 =145.8334− 2640.0334 = −2494.2 J
ΔS(path 1) = nC p ln(250K 450K)− nRln(2 bar 3 bar)
= (1mol× 2.5×8.314 J mol-1K -1)*ln(250 450)
− (1mol ×8.314 J mol-1K -1)*ln(2bar 3 bar)
= −8.85 J K -1
ΔS(path 2) = −nRln(2bar 3bar)+ nC p ln(250K 450K)
= −(1mol ×8.314 J mol-1K -1)*ln(2bar 3bar)
+ (1mol × 2.5×8.314 J mol-1K -1)*ln(250 K 450 K)
= −8.85 J K -1

Loading page 11...

Chapter 3 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
2
2
2. For a heat engine to have a good efficiency there must be a significant difference in the hot and
cold baths that it utilizes. Temperatures are very uniform on the surface of the earth so that
organisms have not been exposed to simultaneous temperature differences in their environment
thus there can be no evolutionary pressure for a heat engine to come about. Of course, extremely
high temperatures would be needed for the hot bath in order to achieve a good efficiency and it is
unlikely that any stable molecules of life could evolve or be stable at such high temperatures.
3. a) kJ3.133(0.75)kJ)(100;75.0Eff 11 ===−= qqw (Eq. 3.3)
kJ3.33kJ)3.133(kJ)100(; 1331 −=−=−−=−=+ qwqwqq
1q (positive) is heat absorbed by the system at the hot temperature. 3q (negative) is heat
discharged by the system at the low temperature. w (negative) is the work done by the
system.
b) 1131 25.0kJ100 qqqqw +−=−−=+=
kJ3.33andkJ3.133Thus 31 +=−= qq
Heat 3q (positive) is absorbed by the system from a cold reservoir and heat 1q (negative)
is discharged to the hot reservoir when work w (positive) is done on the system. The
engine is acting as a refrigerator.
c) kJ125(0.80)kJ)(100Eff1 ==′−=′ wq
kJ2512510013 −=−=′−′−=′ qwq
d) Net heat absorbed at kJ3.8253.33´33cold =−=+= qqT
Net heat discharged at kJ3.81253.133´11hot −=+−=+= qqT
No net work done. In this hypothetical reversible cycle heat has been transferred from a
cold reservoir to a hot reservoir with no input of work. This cannot happen.
e) Assume an efficiency of 0.6. Then 6.01 =′′′′− qw
For kJ67kJ;167(0.60)kJ)100(kJ,100 31 +=′′−=−=′′+=′′ qqw
Engine (a) could then be used to drive this one in reverse, as a refrigerator. As a result
hot11
cold33
atdischargedheatkJ,34167133
atabsorbedheatkJ,346733
Tqq
Tqq
−=−=′′+
=+−=′′+
No net work done. Again, an impossible result occurs.
4. The organisms in the culture grow by using chemical energy from the nutrient medium.
The gain in entropy from breakdown of complex foodstuffs into much simpler, higher
entropy molecules, more than compensates for the loss in entropy associated with the
“growth” of the organism. Metabolic processes also release heat into the surroundings,
further raising the entropy of the surroundings.

Loading page 12...

Chapter 3 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
3
3
5. a) Δ Smix = Δ SA + Δ SB = −nA Rln( pA 1bar) − nBRln( pB 1bar) from Eq. (3.21)
Now nA + nB =1;therefore XA = nA (nA + nB ) = PA (1bar) and XB = nB (nA + nB ) = pB (1bar)
ΔSm,mix = ΔSmix (nA + nB ) = − XA Rln XA − XB Rln XB
b) ΔSm,mix > 0 , because XA <1 and ln XA < 0 , etc
A positive value for ΔSm, mix means that the process should occur spontaneously in an
isolated system, which is what we expect for the mixing process.
c) At constant temperature, ΔGm,mix = ΔHm,mix −T ΔSm,mix (Eq. 3.32)
ΔHm, mix can be positive, negative or zero. For ideal gases or for the formation of an
ideal liquid solution, ΔHm, mix = 0 . In this case ΔGm, mix = −T ΔSm, mix < 0 , which is what
we expect for the spontaneous mixing process.
6. a) The amount of heat released when O(g)H 2 condenses raises the temperature to100°C .
b) 0on)condensati()( =Δ+Δ=Δ= HTqCHq pp
(0.20 mol)(36.5 J mol-1 K -1 )(5 K) + (n mol)(−40.66 kJ mol−1 ) = 0
n = no. of mols of H 2 O(g) condensed = 8.98 × 10−4 mol
where the heat capacity of steam at 100˚C has been approximated as that at 99.6˚C and 1
bar (table 2.2)
c) ΔS = nC p(g)lnT2
T1
+ ΔS (condensation)
= (0.20 mol)(.0365 kJ K−1 mol−1)ln 373
368 + (8.977 ×10−4 mol)( − 40.66kJ mol−1)
373K
ΔS = 6.60×10−7

Loading page 13...

Chapter 3 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
4
4
8. a) The motor will not work. The ratchet and stick are also hit randomly by gas molecules.
Thus, the stick, intended to only allow clockwise rotation of the wheel (ratchet), can be
lifted due to collisions with gas molecules at the ratchet. At the moment when it is lifted,
there is equal probability that the stick will slip forward or backward one notch due to the
random fluctuations of the gas molecules at the paddle. The result is no net motion.
b) If the ratchet is at a lower temperature than the paddle, the likelihood that the stick will
lift due to fluctuations at the ratchet will be much less (at lower temperature, sufficient
energy for this process is less often achieved.) Thus, the lifting of the stick will be
determined by the gas molecules hitting the paddle, and the stick will work as intended,
only allowing clockwise rotation. In this scenario, the second law is not violated, i.e. heat
is not converted to work at constant temperature.
9. ,0<q (metabolism converts chemical energy into heat)
0=ΔT (thermostat); 0≅w (depending on whether gases are produced or consumed)
ΔU ≅ q < 0; ΔH ≅ ΔU < 0 ; 0<ΔG (spontaneous process at constant T, p)
10. a) 1kg H 2 = (1000g) (2.016g mol−1) = 496.0mol
b) 1kg of n-octane = (1000)/(114.23) = 8.754mol octane
Δf H°(H2O,1) = ( − 285.83)(496.0) = −141,780kJ
ΔS° = [S°(H 2O,l)− S°(H 2 , g)− 0.5S°(O 2 ,g)](496 mol)
= [69.91−130.57 − (0.5)(205.03)](496) = −80.94kJ K−1
Δf G° = (−237.18)(496) = −117,650 kJ
12.5 O 2 (g)+ C8 H18 (g)→ 8CO 2 (g)+ 9H 2O(g)
ΔH° = [8(−393.51)+ 9(−285.83)− (−208.45)](8.754) = −48,250 kJ
ΔS° = [8(213.64)+ 9(69.91)− 466.73− (12.5)(205.03)](8.75) = −6.05 kJ K−1
ΔG° = [8(−394.36)+ 9(−237.18)−16.40](8.754) = −46,450 kJ
11. a) 6C(s, graphite) + 3H 2 (g) → C 6 H 6 (g)
ΔrG298
° = Δf G298
° (benzene) =129.86 kJ mol−1
Don’t buy: non-spontaneous at C25° ,1 bar. Catalysts change rate, not spontaneity.
b) 2NO(g)+ O 2 (g)→ 2NO 2 (g)
ΔrG298
° = −2(86.55)+ 0+ 2(51.47) = −70.16 kJ mol−1
Yes, the reaction is spontaneous.
c) 3CH 4 (g)

Loading page 14...

Chapter 3 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
5
5
12. a) 100 g MgATP 2- * (1 mol/527.42 g) (-50 kJ / mol) = -9.48 kJ
b) 6000 kJ / -50 kJ| ( 527.416 g / mol) = 63.3 kg MgATP 2- or 633 turnovers of 100 g
c) 6.3 kg or about 14 lbs of MgATP 2- would be needed for this level of storage; this is still
excessive, and would be around 10% of a typical body mass; for comparison bone mass
percentage is around 15% in most adults
13. a) irreversibly
b) the system plus the surroundings. In this case, the surroundings is a correct answer but
the first answer is true for all spontaneous irreversible processes.
c) enthalpy change
d) less than (consider the analogous process of the Carnot cycle); if the net process is a cycle
for the system then entropy of surroundings must have net increase
14. C 2 H 5 OH(l) → C 2 H 6 (g) + 1 2O 2 (g)
Δr H298
° = −(−276.98)+ (−84.0)+ 0 =192.98 kJ mol−1
ΔrG298
° = −(−174.09)+ (−32.08)+ 0 =142.01 kJ mol−1
ΔrU298
° = Δr H298
° − Δ( pV ) ≅ Δr H298
° − (Δngases )RT (assuming ideal gas behavior)
=192.98 kJ mol−1 − (3 2)(8.314 1000)(298) =189.3 kJ mol−1
To obtain ΔH
ο at500
ο C and 1 bar we would need heat capacity values for each of the reactants
and products over that temperature range. To be realistic, we should include the enthalpy of
vaporization of ethanol, which is a gas at 500
ο C and 1 bar.
Δr H773
° = Δr H298
° + (ΔC p)
298
773
∫ dT + ΔvapH°
773(C 2 H 5OH)
15. aragonite)(CaCOcalcite)(CaCO 33 +
a) ΔG298
° = −(−1128.79)+ (−1127.71) = +1.08 kJ mol−1
No. The positive ΔG° says that the process is non-spontaneous.
b) Because the density of aragonite is greater than that of calcite, an increase in pressure
should favor the conversion to aragonite, according to Le Chatelier’s Principle.
c) ΔrG298 ( p)− ΔrG298
° = ΔrV ( p −1 bar), where ΔrG298 ( p) < 0 (Eq. 3.45)

Loading page 15...

Chapter 3 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
6
6
p = −1,050 J mol−1[(100.09 g mol−1)(1 2.930−1 2.710) cm 3g−1]−1
× (83.14 cm 3 bar deg−1 8.314 J deg−1)+1 bar
= 3790 bar
d) ΔS2 9 8
° = 88.70 − 92.88 = −4.18 J K−1 mol −1
Because ΔS° is negative, that means that ΔG° will increase with increasing temperature,
making the conversion even less favorable.
16. CH 3 COCOOH(l)→ CH 3 CHO(g)+ CO 2 (g)
a) ΔrG298
° = −(−463.33)+ (−133.27)+ (−394.36) = −64.30 kJ mol−1
b) Assuming that both CHOCH 3 and 2CO remain gases at C25° and 100 bar.
ΔG = ΔG°+ RT ln([ p2 (CH 3CHO) p1][ p2 (CO 2 ) p1])
= −64.30+ (8.314 1000)(298) ln (100×100) = −41.48 kJ mol−1
17. a) decrease
b) remain unchanged
c) zero
d) entropy change of the surroundings, entropy change of the universe
e) decrease
f) remain unchanged
g) remain unchanged
h) more negative than
18. a) w = − pφ(Δ φVm )
Δ φU = qφ + w; Δ φH = qφ
Δ φS = qφ
Tφ
; Δ φG = 0
b) Δ φH = qφ + (C p,m,
β − C p,m,
α )(T∗ −Tm)
Δ φS = qφ
Tφ
+ (C p,m,
β − C p,m,
α )lnT∗
Tφ

Loading page 16...

Solution Manual for Precalculus , 10th Edition

Study Now!

Document Details

Related Documents

Test Bank For Precalculus, 11th Edition

Solution Manual for Precalculus Enhanced with Graphing Utilities, MyLab Math, 8th Edition

Solution Manual for Precalculus Enhanced with Graphing Utilities, 8th Edition

Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual

Solution Manual for Precalculus, 11th Edition

Solution Manual for Precalculus, 8th Edition

Test Bank for Precalculus , 10th Edition

Test Bank for Precalculus Enhanced with Graphing Utilities , 7th Edition

Test Bank for Precalculus Enhanced with Graphing Utilities, 6th Edition

AP Calculus AB Scoring Guide Unit 6 Progress Check

Company

Explore

Study Tools