Solution Manual for Precalculus , 10th Edition

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CHAPTER 21.a)qp=ΔH= 1 L×0.9982 kg L–1×2447 kJ kg–1= 2443 kJb)ΔT= 2443 kJ/(60 kg×4.184 kJ K–1kg–1) = 9.7 Kc)C12H22O11(s) + 12O2(g)→12 CO2(g) + 11H2O (l)ΔH= 11(–285.83 kJ mol–1) + 12(–393.51 kJ mol–1) – (–2222.10 kJ mol–1)= –5644 kJ mol–1Msucrose= 342.31 g mol–1msucrose= 2443 kJ×342.31 g mol–1/5644 kJ mol–1= 148.2 g2.a)1 kg of carbon is (1.0 kg×1000 g kg–1)/(12.011 g mol–1) = 83.3 mol83.3 mol CO2×8.3145 J mol–1K–1×298 K/(105Pa) = 2.06 m3CO22.06 m3CO2/(0.000390 m3CO2/m3air ) = 5292 m3airb)p =F/A=mg/A =105Pa.m=Ap/g= 1 m2×105Pa/(9.80662 m s–2) = 10197 kgMair= 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 airnCO2= 354069 mol air×0.000390 mol CO2/(mol air) = 138.09 mol CO2mC= 0.012011 kg/mol C×138.09 mol CO2= 1.659 kg Cc)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 Jb)k=F/(x –x0) = 5.00 N/(0.105 m – 0.100 m) = 1.00×103N m–1w= (k/2)(x–x0)2= 1.00×103N m–1×(0.01 m)2/2 = 0.050 JΔd/dsurface×100% = 1.2%c)w= –p(V2–V1) = –105Pa×0.002 m3= –200 J.d)w= –2×10–4Je)w= –nRTln(V2/V1) = –p1V1ln(V2/V1) = –105Pa×10–3m3×ln 3 = –110 J.4.a)w=f dx=010∫(20x+x2)dx=010∫10x2+13x3010=(1000+333)×10−21J = 1.333×10−18Jb)w=1.333×10−18J×6.022×1023mol–1=8.03×105J mol–15.a)p =ρgh= 1025 kg m–3×9.80662 m s–2×2500 m = 2.513×107Pa = 251.3 barb)ΔV/V= –κΔp= –(2.513×107Pa)×4.9458×10−10Pa−1= –0.0124So the densityincreasesby a factor of 0.0124, or 1.24 %c)p1V1/T1=p2V2/T2, soV2=p1V1T2/(T1p2) = 1 bar×10 L×277 K/(293 K×252.3 bar)

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Chapter 2 solutions manual to accompany 5thedition ofPhysical ChemistryTinoco, Sauer, Wang, Puglisi Harbison, Rovnyak2= 0.0375 L. Density is reciprocally related to volume. 10/0.0375 = 266.9 – 1 = 265.9 =26590%6.nH2O= 100 g/(18.015 g mol–1) = 5.55 mol.a)q= 5.55 mol×75.4 J mol–1K–1×100 K = 4.19×104Jb)q=nH2OΔfusH= 5.55 mol×6007 J mol–1= –3.33×104Jc)q=nH2OΔvapH= 5.55 mol×40667 J mol–1= 2.26×105J7.a)V1= 1 mol×8.3145 J mol–1K–1×300 K /(105Pa) = 0.0249 m3V2= 1 mol×8.3145 J mol–1K–1×600 K /(105Pa) = 0.0499 m3wp= –p(V2–V1) = –105Pa×(0.0499 – 0.0249)m3= –2494 Jb)ΔU=nCV,mΔT= 1 mol×20.8 J mol–1K–1×300 K = 6.24 kJΔH=ΔU+Δ(pV) = 6.24 kJ + 2.49 kJ = 8.73 kJc)qp=ΔH= 8.73 kJ8.a)The volume,V= (4/3)πr3= 1.333×3.142×(5×10–7m)3= 5.236×10–19 m3m=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 = 1.752×1010d)6378.1×103 m/(1×10–6m) = 6.3781×1012e)A =4πr2= 4×3.142×(5×10–7m)2= 3.142×10–12 m2f)3.142×10–12 m2/(0.50×10–18 m2) = 6.283×1069.a)ΔU=ΔH= 0.wT=p(V2–V1) =p×nRT(1/p2– 1/p1)= 105Pa×(1 mol)(8.3145 J K–1mol–1)(298 K)[1/(105 Pa) – 1/(106 Pa)]= 2230 JqT= –wTb)q= 1 mol×(5/2)×8.3145 J mol–1 K–1×75 K = 1559 Jw= 0ΔU=qΔH=ΔU + nRΔT= 1559 J + 1 mol×8.3145 J mol–1 K–1×75 K = 2183 Jc)p= 1 bar×373 K /(298 K) = 1.252 bard)lower, because the system will do work against the surroundings.10.a)q= 0, adiabatic.w< 0, expansion.ΔU=w< 0.ΔH< 0.b)ΔU=ΔH= 0 sinceΔT= 0

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Chapter 2 solutions manual to accompany 5thedition ofPhysical ChemistryTinoco, Sauer, Wang, Puglisi Harbison, Rovnyak3w< 0, expansionq= –w> 0c)q= 0, adiabatic.w= 0, expansion against zero pressureΔU=w+q= 0.ΔH=ΔU+RΔT= 0d)q> 0, heat of vaporization.w< 0, expansion against pressureΔH=qp> 0.ΔU=ΔH–Δ(pV) > 0;ΔH >>Δ(pV)≅ngRTe)q< 0, exothermic reaction.w= 0, closed bomb,ΔV= 0ΔU=qV< 0.ΔH=qV+Δ(pV) < 0;Δ(pV)=–(nH2+nO2)RT< 011.a)ΔH=qp= 40.667 kJ.ΔU=ΔH–Δ(pV) =ΔH–Δ(ngRT) = 40667 J – (1 mol×8.3145 J mol–1K–1×373.15 K)= 37.564 kJwp=pΔV=Δ(ngRT) = 1 mol×8.3145 J mol–1K–1×373.15 K = –3.103 kJb)w=w1+w2= 0, sincepex= 0 in step 1 andΔV= 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−1mol−1)(20 K)=760 Jb)q=(1 mol)(6007 J mol−1)=6007 Jc)Per degree temperature drop, 10 mol H2O (l) loses(10 mol)(75.4 J mol−1K−1)=754 J K−1760 J+6007 J+(1 mol)(75.4 J K−1mol−1)(Tf−0)=(10 mol)(75.4 J K−1mol−1)(20−Tf)Tf=(10)(75.4)(20)−760−6007(11)(75.4)=10.0°CThe final state is liquid.13.a)ΔT=qPnCp,m=−400 J (36.5 J K−1mol−1×1 mol)=–11.0 KTf=125.0−11.0=114.0°CThere is no phase change, soΔV=nRΔT/p=1 mol(8.314 J K−1mol−1)(−11.0 K)105Pa=−9.1×10−4m3

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Chapter 2 solutions manual to accompany 5thedition ofPhysical ChemistryTinoco, Sauer, Wang, Puglisi Harbison, Rovnyak4b)ΔT=−400 J (1 mol×75.95J mol−1K−1)=−5.27 °CTf=100−5.27=94.73°CVf(l)=(18 g)/(0.96 g mL−1)=18.75 mL. No phase change;∆Vis negligible for theliquid water.c)Δng=−400 J()40,657 J mol−1()=−9.84×10−3mol∆ngmol water vapor will condense to liquidNo temperature changeΔV=ΔnRT/p=(−9.84×10−3mol)(8.314 J mol-1K-1)(373 K) / (105Pa)ΔV=−3.05×10−4m3d)Part (a), because it has the greatestΔV;w> 0.14.V= 5.8 – 2×exp[100(0.002)] =3.36 LAt 1.005 bar,V= 5.8 – 2×exp[100(–0.005)] =4.59 LΔn= 105Pa×(4.59 – 3.36)×10–3m3/(8.3145 J mol–1K–1×310 K) = 0.0477 mol15.a)q= 0, thermally insulatedw= 0, assuming no volume change because decrease in volume of hot copperequals increase in cool copper.ΔU= 0, energy is conserved.ΔH= 0b)q= 0, thermally insulatedw= 0, mechanical work done on the systemΔU=w> 0ΔH= 0 becauseΔpV= 0, with negligible expansion of the system.c)q= 0w= 0ΔU= 0ΔH= 0 (all answers assume ideal gas behavior)16.a)First Law of Thermodynamics – no restrictionsb)Constant pressure process with no non-pVwork.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, whereΔ(pV) =RTΔne)van der Waals gas.f)Constantpex.

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Chapter 2 solutions manual to accompany 5thedition ofPhysical ChemistryTinoco, Sauer, Wang, Puglisi Harbison, Rovnyak517.c=Cp,m/M= 75.4 J mol–1K–1/(0.018 kg mol–1) = 4189 J/K.So 4 kJ/hr corresponds to almost 1° K per hour. It’s probably best to limit moon walks totwo hours, since 2°C (3.6 F) above normal is a substantial fever.18.n=pVRT=105Pa()5×10−4m3()8.314 J K−1mol−1()293 K()=0.021 molPer breath,ΔH=nCp,m(T2–T1) = 0.021 mol×3.5×8.3145 J mol–1K–1×17.15 K = 10.2 J.Per day, 10.2 J×12 min–1×60 min hr–1×24 hr d–1= 177 kJ d–1, representing about 1.5% offood energy.Metabolic heat = 80 kg×4 kJ h–1kg–1×24 h = 7.68 kJ d–1, or 0.06% of food energy. At outsidetemperatures of –40°C, 4.5×more energy will be lost through breathing; 7% of food energybegins to be significant.19.a)w=pΔV= 105Pa×5×10–4m3= 50 J/breath.In 24 hr,w= 50 J×15000 d–1= 750 kJb)w=mgh= 750 kJ som=w/gh= 750 kJ/(9.8 m s–2×100 m) = 765 kJ.20.By convention, lower case symbols are specific (per kilogram) rather than molar quantitiesmwatercp,m,waterTf−55()K+micecp,m,ice×10 K+mice()Δfush+micecp,m,waterTf– 0()K=00.100 kg×4.184 kJ K−1kg−1×Tf−55()K+0.010 kg×2.11 kJ K−1kg−110 K()+0.010 kg×333.4 kJ kg−1+0.010 kg×4.184 kJ K−1kg−1×TfK = 00.4184Tf−23.01+0.211+3.334+0.04184Tf=0Tf=19.470.460=42.3°C21.ΔH298°=6ΔfH°CO2,g()+6ΔfH°H2O,l()− ΔfH°C6H12O6,s()−6ΔfH°O2,g()=[6−393.509()+6−285.830()− −1274.4()−0] kJ mol−1=−2801.6 kJ mol−122.a)qp=ΔH=2ΔfH°ethanol,l()+2ΔfH°CO2,g()− ΔfH°glucose,s()=2−276.98()+2−393.509()− −1274.4()⎡⎣⎤⎦kJ mol−1=−66.6kJb)This represents 2.4% of the heat available from the complete combustion.23.ΔrH° =ΔfH°H2O,l()+12ΔfH°O2,g()− ΔfH°H2O2,aq()

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Chapter 2 solutions manual to accompany 5thedition ofPhysical ChemistryTinoco, Sauer, Wang, Puglisi Harbison, Rovnyak6ΔrH° =(−285.830)+0−(−191.17)⎡⎣⎤⎦kJ mol−1=−94.66 kJ mol−1A temperature rise of 0.02 ˚C requires (0.02 ˚C)(4.184 kJ kg–1)(1 kg L–1) = 0.0837 kJ L–1Minimum detectableconcentration=0.0837 kJ L−194.66kJ mol−1=8.84×10−4M24.a)ΔH°298=−74.81()+0− −238.57()⎡⎣⎤⎦kJ mol−1=163.76 kJ mol−1b)ΔU° =ΔH°− ΔngasesRT=163.76 kJ mol−1−328.3145 J mol−1K−11000()298 K()=160.04 kJ mol−1c)ΔH°773=ΔH°298+ΔCpdT298773∫whereΔCp=Cp,mCH4()+12Cp,mO2()−Cp,mCH3OH()25.ΔrH°298=ΔfH°(urea,s)+3ΔfH°(CO2,g)+3ΔfH°(H2O,l)−2ΔfH°(glycine,s)ΔrH°298=−333.17()+3−393.509()+3−285.83()−2−537.2()⎡⎣⎤⎦kJ mol−11molkJ8.1296−−=26a)ΔrH°298=2−276.98()+2−393.509()− −1274.4()⎡⎣⎤⎦kJ mol−1=−66.6 kJ mol−1b)ΔrH°298=2−484.1()+2−393.509()+2−285.830()− −1274.4()⎡⎣⎤⎦kJ mol−11molkJ5.1052−−=c)ΔH298°=−2801.6 kJ mol−1(see Prob. 21)27.a)From problem 26 a)ΔrH°353=ΔrH°298+ΔTΔrCp=–66.6 kJ mol–1+(353 – 298)K(2×111.5+2×37.1–219.2)J mol–1K= –62.3 kJ mol–1b)From problem 26 b)ΔrH°353=–1052.5 kJ mol–1+(353 – 298)K(2×123.5+2×75.4+2×37.1–219.2–2×29.4)J mol–1K= –1041.8 kJ mol–1

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Chapter 2 solutions manual to accompany 5thedition ofPhysical ChemistryTinoco, Sauer, Wang, Puglisi Harbison, Rovnyak728.a)ΔrH°298=ΔfH°sucrose,s()+12ΔfH°O2,g()−12ΔfH°CO2,g()−11ΔfH°H2O,l()=−2222.1()+0−12−393.509()−11−285.830()⎡⎣⎤⎦kJ mol−1=5644.1 kJ mol−1b)20 kgsucrose ha−1hr−1=5644.1 kJ mol−1()0.3423 kg mol−1()−120 kg()=329,800 kJ h−1=91.6 kW ha−1c)Stored energy=91.6 kW hectare−1()10−4hectare m−2()1 kW m−2[]100%92.0=29.a)H2g()+12O2g()→H2Ol( )ΔrH°298=ΔfH°H2O,l()=−285.830 kJ mol−1()2.016 g H2mol−1()=−141.8 kJ g H2()−1b)C8H18g()+252O2g()→8CO2g()+9H2Ol( )ΔrH298°=− −208.45()−0+8−393.509()+9−285.830()⎡⎣⎤⎦114 gC8H18mol()=−48.35 kJ gC8H18()−1c)ΔrH298°H2()ΔrH298°C8H18()=−141.8−48.35=2.9Thus H2is nearly 3 times more energetic as a fuel than octane on a weight basis.30.a)ΔrH° =ΔfH°Ala()+ΔfH°CO2()− ΔfH°Asp()=−562.7()+−393.509()− −973.37()⎡⎣⎤⎦kJ mol−1=17.2kJ mol−1, absorbedHeat capacities neededΔrH323°=ΔrH298°+ΔrCp,m50−25()whereΔrCp,m=Cp,mAla()+Cp,mCO2()−Cp,mAsp()

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Chapter 2 solutions manual to accompany 5thedition ofPhysical ChemistryTinoco, Sauer, Wang, Puglisi Harbison, Rovnyak831.To measureΔrH298°the reactants could be put into a calorimeter at 25 ˚C and a small amount ofan enzyme catalyst added. The heat evolved per mole at constant pressure isΔrH298°To estimateΔrU298°the volume change of the reaction must be determined. The change in volume can bemeasured in a dilatometer, or partial molal volumes can be used to calculate the change involume. For practical purposes the volume change is negligible, since no gases are involved, andΔrU298°=ΔrH298°.32.a)22 kWhd−1()3600 kJ (kWh)−1()=7.92×104kJ d−1b)1 kW m−2()0.1()5 h()=0.5 kWh m−2A=22 kWh()0.5 kWh m−2()=44 m233.a)ΔrH298°=2ΔfH°H2O,g()+ΔfH°O2,g()− ΔfH°H2O2,g()=2−241.818()+0−2−133.18()⎡⎣⎤⎦kJ mol−1=−217.28 kJ mol−1b)()()()1molkJ5.1403.4982128.21721OOD−=+−=−c)ΔrH°298=2−285.830()+0−2−191.17()⎡⎣⎤⎦kJ mol−1=−189.32 kJ mol−1d)Heat evolved=1 2189.32 kJ (mol H2O2)−1()0.01 molkg−1()=0.947 kJ kg−1Temperature rise=0.947 kJ kg−1()4.18 kJ K−1kg−1()=0.226 KFinal temperature25.226°C34.a)8 C graphite()+9H2g()→C8H18g()ΔfH° =8 716.7()+9 436.0()−7 359()−18 411()⎡⎣⎤⎦kJ mol−1=−253 kJ mol−1Published value = –208.45 kJ mol–1

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Chapter 2 solutions manual to accompany 5thedition ofPhysical ChemistryTinoco, Sauer, Wang, Puglisi Harbison, Rovnyak9Chief discrepancy lies in using average bond dissociation energies for D(C—C) andD(C—H)b) 10 C (graphite) + 4 H2(g)→ΔfH° =10ΔfH°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−1Published value1molkJ96.150−Chief discrepancy arises from neglect of resonance energy.c)C graphite()+H2g()+12O2g()→H2C=Og()ΔfH° =ΔfH°C()+D H−H()+12D O2()−2D C−H()−D C=O()=716.7+436.0+498.3 / 2−2 411()−709=−129 kJ mol−1Published value1molkJ90.115−−Chief discrepancy arises from use of average bond dissociation energies.d)C (graphite) + H2(g) + O2(g)→formic acidΔHf°=ΔHfC()+D H−H()+D O2()−D C−H()−D C=O()−D C−O()−D O−H()=716.7+436.0+498.3−411−709−361−452=−282 kJ mol−1Published value−424.76+46.15=−378.61 kJ mol−1Chief discrepancy comes from neglect of hydrogen bonding.35.The steric repulsion can be estimated by the enthalpy change for the processcis-2-butene→trans-2-buteneΔrH°298=−11.1 kJ mol−1− −7.0 kJ mol−1()=−4.1 kJ mol−1by which we see that the trans form has the lower enthalpy by 4.1 kJ/mol.

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3-14CHAPTER 31.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 reversiblepath 2: the first step is isothermal, the second is isobaric; both reversiblework(path 1) =-nR(250-450K)+nR(250K)ln(2bar 3bar)=-(1mol×8.314J mol−1K−1)(−200K)+(1mol×8.314J mol−1K−1)(250K)ln(2bar 3bar)=820.0408 Jq(path 1)=nCp(250−450K)−nR(250K)ln(2bar 3bar)=(1mol×2.5*8.314J mol−1K−1)(−200K)−(1mol×8.314J mol−1K−1)(250)ln(2 3)=−3314.2408 Jwork(path 2) =nR(450K)ln(2bar 3bar)-nR(250-450K)=(1mol×8.314J mol−1K−1)(450K)ln(2bar 3bar)−(1mol×8.314J mol−1K−1)(−200)=145.8334 Jq(path 2)=−nR(450K)ln(2bar 3bar)+nCp(250−450K)=−(1mol×8.314J mol−1K−1)(450K)ln(2bar 3bar)+(1mol×2.5×8.314J mol−1K−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=−2494.2J(first law)ΔU(path 2)=qpath 2+wpath 2=145.8334−2640.0334=−2494.2JΔS(path 1)=nCpln(250K 450K)−nRln(2 bar 3bar)=(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-1ΔS(path 2)=−nRln(2bar 3bar)+nCpln(250K 450K)=−(1mol×8.314 J mol-1K-1)*ln(2bar 3bar)+(1mol×2.5×8.314 J mol-1K-1)*ln(250 K 450K)=−8.85 J K-1

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Chapter 3 solutions manual to accompany 5’th Edition ofPhysical Chemistry,Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak222.For a heat engine to have a good efficiency there must be a significant difference in the hot andcold baths that it utilizes. Temperatures are very uniform on the surface of the earth so thatorganisms have not been exposed to simultaneous temperature differences in their environmentthus there can be no evolutionary pressure for a heat engine to come about. Of course, extremelyhigh temperatures would be needed for the hot bath in order to achieve a good efficiency and it isunlikely that any stable molecules of life could evolve or be stable at such high temperatures.3.a)kJ3.133(0.75)kJ)(100;75.0Eff11===−=qqw(Eq. 3.3)kJ3.33kJ)3.133(kJ)100(;1331−=−=−−=−=+qwqwqq1q(positive) is heat absorbed by the system at the hot temperature.3q(negative) is heatdischarged by the system at the low temperature. w (negative) is the work done by thesystem.b)113125.0kJ100qqqqw+−=−−=+=kJ3.33andkJ3.133Thus31+=−=qqHeat3q(positive) is absorbed by the system from a cold reservoir and heat1q(negative)is discharged to the hot reservoir when work w (positive) is done on the system. Theengine is acting as a refrigerator.c)kJ125(0.80)kJ)(100Eff1==′−=′wqkJ2512510013−=−=′−′−=′qwqd)Net heat absorbed atkJ3.8253.33´33cold=−=+=qqTNet heat discharged atkJ3.81253.133´11hot−=+−=+=qqTNo net work done. In this hypothetical reversible cycle heat has been transferred from acold reservoir to a hot reservoir with no input of work. This cannot happen.e)Assume an efficiency of 0.6. Then6.01=′′′′−qwForkJ67kJ;167(0.60)kJ)100(kJ,10031+=′′−=−=′′+=′′qqwEngine (a) could then be used to drive this one in reverse, as a refrigerator. As a resulthot11cold33atdischargedheatkJ,34167133atabsorbedheatkJ,346733TqqTqq−=−=′′+=+−=′′+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, higherentropy 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.

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Chapter 3 solutions manual to accompany 5’th Edition ofPhysical Chemistry,Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak335.a)ΔSmix=ΔSA+ΔSB=−nARln(pA1bar)−nBRln(pB1bar)from Eq. (3.21)NownA+nB=1;thereforeXA=nA(nA+nB)=PA(1bar) andXB=nB(nA+nB)=pB(1bar)ΔSm,mix=ΔSmix(nA+nB)=−XARlnXA−XBRlnXBb)ΔSm,mix>0, becauseXA<1andlnXA<0, etcA positive value forΔSm, mixmeans that the process should occur spontaneously in anisolated system, which is what we expect for the mixing process.c)At constant temperature,ΔGm,mix=ΔHm,mix−TΔSm,mix(Eq. 3.32)ΔHm, mixcan be positive, negative or zero. For ideal gases or for the formation of anideal liquid solution,ΔHm, mix=0. In this caseΔGm, mix=−TΔSm, mix<0, which is whatwe expect for the spontaneous mixing process.6.a)The amount of heat released whenO(g)H2condenses raises the temperature to100°C.b)0on)condensati()(=Δ+Δ=Δ=HTqCHqpp(0.20 mol)(36.5 J mol-1K-1)(5 K) + (nmol)(−40.66 kJ mol−1) = 0n= no. of mols of H2O(g) condensed = 8.98×10−4molwhere the heat capacity of steam at 100˚C has been approximated as that at 99.6˚C and 1bar (table 2.2)c)ΔS=nCp(g)lnT2T1+ΔS(condensation)=(0.20 mol)(.0365 kJ K−1mol−1)ln 373368+(8.977×10−4mol)(−40.66kJ mol−1)373KΔS=6.60×10−7kJ K−17.(a)This is a statement of the Second Law. If two reversible heat engines have differentefficiencies then they could be arranged in parallel between two heat baths such that heatis spontaneously transferred from the cold bath to the hot bath.(b)This is incorrect and is not a statement of the Second Law. Rather, a version of theSecond Law as put forward by Kelvin (W. Thomson) is that no engine operating in acycle can accept a given quantity of heat from one source and convert it solely to work.(c)This is a statement of the second law, although this concept is usually stated as theimpossibility of the reverse direction put forward by Clausius that no engine working in acycle can produce as its only result the transfer of heat from a cold bath to a hot bath.(d)This is incorrect and is not a statement of the Second Law. Rather the entropy of theuniverse is increasing.

Solution Manual for Precalculus , 10th Edition - Page 14 preview image

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Chapter 3 solutions manual to accompany 5’th Edition ofPhysical Chemistry,Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak448.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 belifted 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 therandom 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 willlift due to fluctuations at the ratchet will be much less (at lower temperature, sufficientenergy for this process is less often achieved.) Thus, the lifting of the stick will bedetermined 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. heatis 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 constantT, p)10.a)1kg H2=(1000g) (2.016g mol−1)=496.0molb)1kg ofn-octane=(1000)/(114.23)=8.754mol octaneΔfH°(H2O,1)=(−285.83)(496.0)=−141,780kJΔS° =[S°(H2O,l)−S°(H2, g)−0.5S°(O2,g)](496 mol)=[69.91−130.57−(0.5)(205.03)](496)=−80.94kJ K−1ΔfG° =(−237.18)(496)=−117,650 kJ12.5 O2(g)+C8H18(g)→8CO2(g)+9H2O(g)ΔH° =[8(−393.51)+9(−285.83)−(−208.45)](8.754)=−48,250 kJΔS° =[8(213.64)+9(69.91)−466.73−(12.5)(205.03)](8.75)=−6.05 kJ K−1ΔG° =[8(−394.36)+9(−237.18)−16.40](8.754)=−46,450 kJ11.a)6C(s, graphite)+3H2(g)→C6H6(g)ΔrG298°=ΔfG298°(benzene)=129.86 kJ mol−1Don’t buy: non-spontaneous atC25°,1 bar. Catalysts change rate, not spontaneity.b)2NO(g)+O2(g)→2NO2(g)ΔrG298°=−2(86.55)+0+2(51.47)=−70.16 kJ mol−1Yes, the reaction is spontaneous.c)3CH4(g)+NH3(g)+3O2(g)→CH3CHNH2COOH(s)+4H2O(l)ΔrG298°=−3(−50.70)−(−16.78)+0+(−371.64)+4(−237.13)=−1151.28 kJ mol−1Yes, this reaction is spontaneous.

Solution Manual for Precalculus , 10th Edition - Page 15 preview image

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Chapter 3 solutions manual to accompany 5’th Edition ofPhysical Chemistry,Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak5512.a)100 g MgATP2-* (1 mol/527.42 g) (-50 kJ / mol) = -9.48 kJb)6000 kJ / -50 kJ| ( 527.416 g / mol) = 63.3 kg MgATP2-or 633 turnovers of 100 gc)6.3 kg or about 14 lbs of MgATP2-would be needed for this level of storage; this is stillexcessive, and would be around 10% of a typical body mass; for comparison bone masspercentage is around 15% in most adults13.a)irreversiblyb)the system plus the surroundings. In this case, the surroundings is a correct answer butthe first answer is true for all spontaneous irreversible processes.c)enthalpy changed)less than (consider the analogous process of the Carnot cycle); if the net process is a cyclefor the system then entropy of surroundings must have net increase14.C2H5OH(l)→C2H6(g)+1 2O2(g)ΔrH298°=−(−276.98)+(−84.0)+0=192.98 kJ mol−1ΔrG298°=−(−174.09)+(−32.08)+0=142.01 kJ mol−1ΔrU298°=ΔrH298°− Δ(pV)≅ ΔrH298°−(Δngases)RT(assuming ideal gas behavior)=192.98 kJ mol−1−(3 2)(8.314 1000)(298)=189.3 kJ mol−1To obtainΔHοat500οCand 1 bar we would need heat capacity values for each of the reactantsand products over that temperature range. To be realistic, we should include the enthalpy ofvaporization of ethanol, which is a gas at500οCand 1 bar.ΔrH773°=ΔrH298°+(ΔCp)298773∫dT+ΔvapH°773(C2H5OH)15.aragonite)(CaCOcalcite)(CaCO33+a)ΔG298°=−(−1128.79)+(−1127.71)= +1.08 kJ mol−1No. The positiveΔG°says that the process is non-spontaneous.b)Because the density of aragonite is greater than that of calcite, an increase in pressureshould favor the conversion to aragonite, according to Le Chatelier’s Principle.c)ΔrG298(p)− ΔrG298°=ΔrV(p−1 bar), whereΔrG298(p)<0(Eq. 3.45)

Solution Manual for Precalculus , 10th Edition - Page 16 preview image

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Chapter 3 solutions manual to accompany 5’th Edition ofPhysical Chemistry,Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak66p=−1,050 J mol−1[(100.09 g mol−1)(1 2.930−1 2.710) cm3g−1]−1×(83.14 cm3bar deg−18.314J deg−1)+1 bar=3790 bard)ΔS2 9 8°=88.70−92.88=−4.18 J K−1mol−1BecauseΔS°is negative, that means thatΔG°will increase with increasing temperature,making the conversion even less favorable.16.CH3COCOOH(l)→CH3CHO(g)+CO2(g)a)ΔrG298°=−(−463.33)+(−133.27)+(−394.36)=−64.30 kJ mol−1b)Assuming that bothCHOCH3and2COremain gases atC25°and 100 bar.ΔG=ΔG°+RTln([p2(CH3CHO)p1][p2(CO2)p1])=−64.30+(8.314 1000)(298) ln (100×100)=−41.48 kJ mol−117.a)decreaseb)remain unchangedc)zerod)entropy change of the surroundings, entropy change of the universee)decreasef)remain unchangedg)remain unchangedh)more negative than18.a)w=−pφ(ΔφVm)ΔφU=qφ+w;ΔφH=qφΔφS=qφTφ;ΔφG=0b)ΔφH=qφ+(Cp,m,β−Cp,m,α)(T∗−Tm)ΔφS=qφTφ+(Cp,m,β−Cp,m,α)lnT∗Tφ

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