CHAPTER 19 SPONTANEOUS CHANGE: ENTROPY AND GIBBS ENERGY

CHAPTER 9 SPONTANEOUS CHANGE: ENTROPY AND GIBBS ENERGY PRACTICE EXAMPLES A (E) In general, S 0 if n gas 0. This is because gases are very disersed cmared t liquids r slids; (gases ssess large entries). Recall that n gas is the difference between the sum f the stichimetric cefficients f the gaseus rducts and a similar sum fr the reactants. (a) n gas =+0+=. One mle f gas is cnsumed here. We redict S 0. (b) n gas = + 0 0 = +. Since ne mle f gas is rduced, we redict S 0. B (E) (a) The utcme is uncertain in the reactin between ZnS(s) and Ag Os. We have used n gas t estimate the sign f entry change. There is n gas invlved in this reactin and thus ur redictin is uncertain. (b) In the chlr-alkali rcess the entry increases because tw mles f gas have frmed where nne were riginally resent ( n gas ( 0) (0 0) A (E) Fr a varizatin, G =0= H TS. Thus, S = H / T. va va va va va va We substitute the given values. S va = H va = T va 0. kj ml 9.79 + 7.5 = 8.0 J ml B (E) Fr a hase change, G = 0 = H TS. Thus, H = TS. We substitute in the tr tr tr given values. H tr = T S tr = 95.5 + 7..09 J ml = 40 J/ml tr tr A (M) The entry change fr the reactin is exressed in terms f the standard entries f the reagents. S =S NH g S N g S H g = 9.5 J ml 9.6 J ml 0.7 J ml = 98.7 J ml Thus t frm ne mle f NH g, the standard entry change is 99.4 J ml 946

B (M) The entry change fr the reactin is exressed in terms f the standard entries f the reagents. S = S NO g + S NO g S N O g 8.5 J ml = 0.8 J ml + 40. J ml S NOg = 450.9 J ml S N Og S N O g = 450.9 J ml 8.5 J ml =.4 J ml 4A (E) (a) Because n gas = += fr the synthesis f ammnia, we wuld redict S 0 fr the reactin. We already knw that H 0. Thus, the reactin falls int case, namely, a reactin that is sntaneus at lw temeratures and nnsntaneus at high temeratures. (b) Fr the frmatin f ethylene n gas = +0= and thus S 0. We are given that H 0 and, thus, this reactin crresnds t case 4, namely, a reactin that is nn-sntaneus at all temeratures. 4B (E) (a) Because n gas = + fr the decmsitin f calcium carbnate, we wuld redict S 0 fr the reactin, favring the reactin at high temeratures. High temeratures als favr this endthermic H 0 reactin. (b) The rasting f ZnS(s) has n gas = = and, thus, S 0. We are given that H 0 ; thus, this reactin crresnds t case, namely, a reactin that is sntaneus at lw temeratures, and nn-sntaneus at high nes. 5A (E) The exressin G = H TS is used with T =98.5. G = H T S = 648 kj 98.5 549. J = 648 kj +6.8 kj = 484 kj kj /000 J 5B 6A (M) We just need t substitute values frm Aendix D int the sulied exressin. G = Gf NOg Gf NOgGf O g = 5. kj ml 86.55 kj ml 0.00 kj ml = 70.48 kj ml (M) Pressures f gases and mlarities f slutes in aqueus slutin aear in thermdynamic equilibrium cnstant exressins. Pure slids and liquids (including slvents) d nt aear. (a) = P SiCl 4 P (b) = HOCl H + Cl P Cl Cl = fr (a) because all terms in the exressin are gas ressures. 947

6B (M) We need the balanced chemical equatin in rder t write the equilibrium cnstant exressin. We start by translating names int frmulas. PbS+HNO s aq PbNO aq+ss+nog The equatin then is balanced with the in-electrn methd. + + xidatin :{PbS s Pb aq +S s + e } reductin :{NO aq + 4H aq + e NO g + H O(l) } + + net inic : PbS s + NO aq +8H aq Pb aq + S s + NO g + 4H O(l) In writing the thermdynamic equilibrium cnstant, recall that neither ure slids (PbS(s) and S(s)) nr ure liquids HO(l) aear in the thermdynamic equilibrium cnstant exressin. Nte als that we have written H + aq here fr brevity even thugh we understand that H O + aq is the acidic secies in aqueus slutin. [Pb ] NO [NO ] [H ] 8 7A 7B (M) Since the reactin is taking lace at 98.5, we can use standard free energies f frmatin t calculate the standard free energy change fr the reactin: NO(g) NO(g) 4 G = Gf NO g Gf N O4 g 5. kj/ml 97.89 kj/ml 4.7kJ G rxn = +4.7 kj. Thus, the frward reactin is nn-sntaneus as written at 98.5. (M) In rder t answer this questin we must calculate the reactin qutient and cmare it t the value fr the reactin: (0.5) NO 4(g) NO (g) Q 0.5 0.5 0.5 bar 0.5 bar G rxn = +4.7 kj = RTln ; 4.7 kj/ml = (8.45 0 kj/ml)(98.5 )ln Therefre, = 0.48. Since Q is greater than, we can cnclude that the reverse reactin will rceed sntaneusly, i.e. NO will sntaneusly cnvert int N O 4. 8A (D) We first determine the value f G and then set G = RTln + G Gf Ag aq + Gf[I (aq)] Gf AgI s [(77.5.57) ( 66.9)] kj/ml 9.7 t determine. 948

8B G 9.7 kj / ml 000 J ln 7.00 RT kj 8.45 J ml 98.5 7.00 7 e 8.50 This is recisely equal t the value fr the s f AgI listed in Aendix D. (D) We begin by translating names int frmulas. MnO +HCl s aq Mn + aq+cl aq equatin with the in-electrn methd. Then we rduce a balanced net inic + + + + xidatin : Cl aq Cl g + e reductin : MnO s + 4H aq + e Mn aq + H O(l) net inic : MnO s + 4H aq + Cl aq Mn aq + Cl g + H O(l) Next we determine the value f G fr the reactin and then the value f. + f f f + Gf MnOs 4 Gf H aq Gf Cl aq 465. kj 4 0.0 kj. kj = +5. kj G = G Mn aq + G Cl g +G H O l = 8. kj + 0.0 kj + 7. kj G ( 5. 0 J ml ) ln 0.7 e 4 0 - RT 8.45 J ml 98.5 0. 5 Because the value f is s much smaller than unity, we d nt exect an areciable frward reactin. 9A (M) We set equal the tw exressins fr G and slve fr the abslute temerature. G = H TS = RTln H = TS RTln = T S Rln H 4.0 J/ml T = = = 607 S Rln 46.4 8.45 ln 50J ml 9B (D) We exect the value f the equilibrium cnstant t increase as the temerature decreases since this is an exthermic reactin and exthermic reactins will have a larger equilibrium cnstant (shift right t frm mre rducts), as the temerature decreases. Thus, we exect t be larger than 000, which is its value at 4. 0. (a) The value f the equilibrium cnstant at 5 C is btained directly frm the value f G, since that value is als fr 5 C. Nte: G = H TS = 77. kj/ml 98.5 0. kj/ml = 40.9 kj/ml G ( 40.90 J ml ) +6.5 7 ln 6.5 e.50 RT 8.45 J ml 98.5 949

(b) First, we slve fr G at 75 C = 48 kj 000 J J G = H TS = 77. 48.5. ml kj ml = 4.870 J/ml Then we use this value t btain the value f the equilibrium cnstant, as in art (a). G ( 4.87 0 J ml ) +.05 5 ln.05 e.7 0 - RT 8.45 J ml 48.5 As exected, fr this exthermic reactin decreases with increasing temerature. 0A (M) We use the value f =9. 0 at 800 and H =.80 5 J/ml, fr the arriate terms, in the van't Hff equatin. 5 5.80.80 J/ml 9.668.45 ln = = 9.66; = 5 9.0 8.45 J ml 800 T T 800.80 4 4 / T =.50 4.50 = 8.00 T = 40 970 C This temerature is an estimate because it is an extralated int beynd the range f the data sulied. 0B (M) The temerature we are cnsidering is 5 C = 508. We substitute the value f =9. 0 at 800 and H =.80 5 J/ml, fr the arriate terms, in the van't Hff equatin. 5.80 J/ml +5. 6 6 ln = =+5.6; =e =6 0 9.0 8.45 J ml 800 508 9.0 6 9 = 6 0 9. 0 = 5 0 INTEGRATIVE EXAMPLE A (D) The value f G can be calculated by finding the value f the equilibrium cnstant at 5 C. The equilibrium cnstant fr the reactin is simly given by {N O 5 (g)}. The var ressure f N O 5 (g) can be determined frm the Clausius-Claeyrn equtin, which is a secialized versin f the van t Hff equatin. Stewise arach: We first determine the value f H sub. ln H sub R T T ln 760 mmhg 00 mmhg H sub 8.4Jml - - 7.5 7.5.4 7.5.08 H sub.49 0 5.8 5 04 J/ml Using the same frmula, we can nw calculate the var ressure f N O 5 at 5 C. 950

ln 00 mmhg 5.8 04 J/ml 8.4Jml - - 80.7 98..46 00 mmhg e.46 4. 4. 00mmHg atm 760 mmhg 0.567atm G RT ln (8.4 0 kjml - - 98.5)ln(0.567).4kJ/ml Cnversin athway arach: ln H sub Rln R T T H sub T T 8.4Jml - - ln 760mmHg 00mmHg.08 H sub 7.5 7.5.49 0 5 Jml- 5.8 0 4 Jml -.4 7.5 ln H sub R 00mmHg e H sub T T e R 5.80 4 Jml - 8.4 J - ml - 80.7 98. - T T 4mmHg atm 760mmHg 0.567atm G RT ln (8.4 0 kjml - - 98.5)ln(0.567).4kJ/ml B (D) The standard entry change fr the reactin ( S ) can be calculated frm the knwn values f H and G. Stewise arach: G H T S S H G 454.8kJml- (.kjml - ) 44.7J - ml - T 98.5 Plausible chemical reactin fr the rductin f ethylene glycl can als be written as: C(s)+H (g)+o (g) CH OHCH OH(l) Since S {S rducts } {S reac tan ts } it fllws that: S rxn S (CH OHCH OH(l)) [ S (C(s)) S (H (g)) S (O (g))] 44.7J - ml - S (CH OHCH OH(l)) [ 5.74J - ml - 0.7J - ml - 05.J - ml - ] S (CH OHCH OH(l)) 44.7J - ml - 608.68J - ml - 67J - ml - 95

Cnversin athway arach: G H T S S H G 454.8kJml- (.kjml - ) 44.7J - ml - T 98.5 44.7J - ml - S (CH OHCH OH(l)) [ 5.74J - ml - 0.7J - ml - 05.J - ml - ] S (CH OHCH OH(l)) 44.7J - ml - 608.68J - ml - 67J - ml - Sntaneus Change and Entry EXERCISES. (E) (a) The freezing f ethanl invlves a decrease in the entry f the system. There is a reductin in mbility and in the number f frms in which their energy can be stred when they leave the slutin and arrange themselves int a crystalline state. (b) (c) The sublimatin f dry ice invlves cnverting a slid that has little mbility int a highly disersed var which has a number f ways in which energy can be stred (rtatinal, translatinal). Thus, the entry f the system increases substantially. The burning f rcket fuel invlves cnverting a liquid fuel int the highly disersed mixture f the gaseus cmbustin rducts. The entry f the system increases substantially.. (E) Althugh there is a substantial change in entry invlved in (a) changing HO (iq., atm) t HO (g, atm), it is nt as large as (c) cnverting the liquid t a gas at 0 mmhg. The gas is mre disersed, (less rdered), at lwer ressures. In (b), if we start with a slid and cnvert it t a gas at the lwer ressure, the entry change shuld be even larger, since a slid is mre rdered (cncentrated) than a liquid. Thus, in rder f increasing S, the rcesses are: (a) (c) (b).. (E) The first law f thermdynamics states that energy is neither created nr destryed (thus, The energy f the universe is cnstant ). A cnsequence f the secnd law f thermdynamics is that entry f the universe increases fr all sntaneus, that is, naturally ccurring, rcesses (and therefre, the entry f the universe increases tward a maximum ). 4. (E) When llutants are rduced they are usually disersed thrughut the envirnment. These llutants thus start in a relatively cmact frm and end u disersed thrughut a large vlume mixed with many ther substances. The llutants are highly disersed, thus, they have a high entry. Returning them t their riginal cmact frm requires reducing this entry, which is a highly nn-sntaneus rcess. If we have had enugh fresight t retain these llutants in a reasnably cmact frm, such as dissing f them in a secure landfill, rather than disersing them in the atmshere r in rivers and seas, the task f ermanently remving them frm the envirnment, and erhas even cnverting them t useful frms, wuld be cnsiderably easier. 95

5. (E) (a) Increase in entry because a gas has been created frm a liquid, a cndensed hase. (b) (c) (d) Decrease in entry as a cndensed hase, a slid, is created frm a slid and a gas. Fr this reactin we cannt be certain f the entry change. Even thugh the number f mles f gas rduced is the same as the number that reacted, we cannt cnclude that the entry change is zer because nt all gases have the same mlar entry. H Sg +O g H Og +SO g Decrease in entry since five mles f gas with high entry becme nly fur mles f gas, with abut the same quantity f entry er mle. 6. (E) (a) At 75 C, ml HO (g, atm) has a greater entry than ml H O (iq., atm) since a gas is much mre disersed than a liquid. (b) ml Fe 50.0 g Fe = 0.896 ml Fe has a higher entry than 0.80 ml Fe, bth (s) 55.8 g Fe at atm and 5 C, because entry is an extensive rerty that deends n the amunt f substance resent. (c) ml Br (iq., atm, 8 C ) has a higher entry than ml Br (s, atm, 8 C) because slids are mre rdered (cncentrated) substances than are liquids, and furthermre, the liquid is at a higher temerature. (d) 0. ml SO (g, 0.0 atm,.5 C ) has a higher entry than 0.84 ml O (g, 5.0 atm,. C ) fr at least three reasns. First, entry is an extensive rerty that deends n the amunt f substance resent (mre mles f SO than O ). Secnd, entry increases with temerature (temerature f SO is greater than that fr O. Third, entry is greater at lwer ressures (the O has a much higher ressure). Furthermre, entry generally is higher er mle fr mre cmlicated mlecules. 7. (E) (a) Negative; A liquid (mderate entry) cmbines with a slid t frm anther slid. (b) (c) (d) (e) Psitive; One mle f high entry gas frms where n gas was resent befre. Psitive; One mle f high entry gas frms where n gas was resent befre. Uncertain; The number f mles f gaseus rducts is the same as the number f mles f gaseus reactants. Negative; Tw mles f gas (and a slid) cmbine t frm just ne mle f gas. 8. (M) The entry f frmatin f a cmund wuld be the difference between the abslute entry f ne mle f the cmund and the sum f the abslute entries f the arriate amunts f the elements cnstituting the cmund, with each secies in its mst stable frm. 95

Stewise arach: It seems as thugh CS wuld have the highest mlar entry f frmatin f the cmunds listed, since it is the nly substance whse frmatin des nt invlve the cnsumtin f high entry gaseus reactants. This redictin can be checked by determining S f values frm the data in Aendix D: (a) Cgrahite + H g CH g S f 4 CH 4 g = S CH 4 g S Cgrahite S H g = 86. J ml 5.74 J ml 0.7 J ml = 80.8 J ml (b) Cgrahite + H g + O g CH CH OH l S f CHCHOH l = S CHCHOH l S Cgrahite S Hg S Og = 60.7J ml 5.74 J ml 0.7 J ml 05.J ml = 45.4 J ml (c) Cgrahite + Srhmbic CS l S CS l = S CS l S Cgrahite S Srhmbic f = 5. J ml 5.74 J ml.80 J ml = 8.0 J ml Cnversin athway arach: CS wuld have the highest mlar entry f frmatin f the cmunds listed, because it is the nly substance whse frmatin des nt invlve the cnsumtin f high entry gaseus reactants. (a) Cgrahite + H g CH g S f 4 CH 4 g = 86. J ml 5.74 J ml 0.7 J ml 80.8 J ml (b) Cgrahite + H g + O g CH CH OH l S f CH CH OH l = (60.7 5.74 0.7 05.)J ml = 45.4 J ml (c) Cgrahite + Srhmbic CS l S f CS l = 5. J ml 5.74 J ml.80 J ml = 8.0 J ml 954

Phase Transitins 9. (M) (a) Hva Hf [HO(g)] Hf [HO(l)] 4.8 kj/ml ( 85.8kJ/ml) 44.0 kj/ml S va = S H O g S H O l = 88.8 J ml 69.9 J ml = 8.9 J ml There is an alternate, but incrrect, methd f btaining S va. S H va 44.00 J/ml va = = = 48 J ml T 98.5 This methd is invalid because the temerature in the denminatr f the equatin must be the temerature at which the liquid-var transitin is at equilibrium. Liquid water and water var at atm ressure (standard state, indicated by ) are in equilibrium nly at 00 C = 7. (b) The reasn why H va is different at 5 C frm its value at 00 C has t d with the heat required t bring the reactants and rducts dwn t 98 frm 7. The secific heat f liquid water is higher than the heat caacity f steam. Thus, mre heat is given ff by lwering the temerature f the liquid water frm 00 C t 5 C than is given ff by lwering the temerature f the same amunt f steam. Anther way t think f this is that hydrgen bnding is mre disruted in water at 00 C than at 5 C (because the mlecules are in raid thermal mtin), and hence, there is nt as much energy needed t cnvert liquid t var (thus H va has a smaller value at 00 C. The reasn why S va has a larger value at 5 C than at 00 C has t d with disersin. A var at atm ressure (the case at bth temeratures) has abut the same entry. On the ther hand, liquid water is mre disrdered (better able t diserse energy) at higher temeratures since mre f the hydrgen bnds are disruted by thermal mtin. (The hydrgen bnds are ttally disruted in the tw vars). 0. (M) In this rblem we are given standard enthalies f the frmatin ( H f ) f liquid and gas entane at 98.5 and asked t estimate the nrmal biling int f entane, G va and furthermre cmment n the significance f the sign f G va. The general strategy in slving this rblem is t first determine H va frm the knwn enthalies f frmatin. Trutn s rule can then be used t determine the nrmal biling int f entane. Lastly, G va,98 can be calculated using G = H TS. va va va 955

Stewise arach: Calculate H va frm the knwn values f H f (art a): C 5 H (l) C 5 H (g) H f -7.5 kjml - -46.9 kjml - H va 46.9 (7.5)kJml - 6.6kJml - Determine nrmal biling int using Trutn s rule (art a): S va H va 87Jml - - T nb T nb H va S va 6.6kJml 87kJ ml 06 000 T nb.9 C Use Gva = Hva TSva t calculate G va,98 (art b): G = H TS va va va G va,98 G va,98 6.6kJml - 98.5 87kJml- - 0.66kJml - 000 Cmment n the value f G va,98 (art c): The sitive value f G va indicates that nrmal biling (having a var ressure f.00 atm) fr entane shuld be nn-sntaneus (will nt ccur) at 98. The var ressure f entane at 98 shuld be less than.00 atm. Cnversin athway arach: C 5 H (l) C 5 H (g) H f -7.5 kjml - -46.9 kjml - H va S va G va 46.9 (7.5)kJml - 6.6kJml - H va T nb = H va 87Jml - - T nb H va S va T S va 6.6kJml- 87kJ - ml - 000 06 6.6kJml - 98.5 87kJml- - 0.66kJml - 000. (M) Trutn's rule is beyed mst clsely by liquids that d nt have a high degree f rder within the liquid. In bth HF and CH OH, hydrgen bnds create cnsiderable rder within the liquid. In C6H5CH, the nly attractive frces are nn-directinal Lndn frces, which have n referred rientatin as hydrgen bnds d. Thus, f the three chices, liquid C6H5CH wuld mst clsely fllw Trutn s rule. 956

. (E) Hva Hf [Br (g)] Hf [Br (l)] 0.9 kj/ml 0.00 kj/ml = 0.9 kj/ml H H S = 87 J ml r = =.50 va va 0.9 0 J/ml va Tva Tva Sva 87 J ml The acceted value f the biling int f brmine is 58.8 C = =.0. Thus, ur estimate is in reasnable agreement with the measured value.. (M) The liquid water-gaseus water equilibrium HO (l, 0.50 atm) HO (g, 0.50 atm) can nly be established at ne temerature, namely the biling int fr water under 0.50 atm external ressure. We can estimate the biling int fr water under 0.50 atm external ressure by using the Clausius-Claeyrn equatin: P ln P = H R va T T We knw that at 7, the ressure f water var is.00 atm. Let's make P =.00 atm, P = 0.50 atm and T = 7. Thus, the biling int under 0.50 atm ressure is T. T find T we simly insert the arriate infrmatin int the Clausius-Claeyrn equatin and slve fr T : 0.50 atm ln.00 atm = 40.7 kj ml 8.45 0 kj ml -.46 0 4 = 7 T 7 T Slving fr T we find a temerature f 54 r 8C. Cnsequently, t achieve an equilibrium between gaseus and liquid water under 0.50 atm ressure, the temerature must be set at 54. 4. (M) Figure -9 (hase diagram fr carbn dixide) shws that at 60C and under atm f external ressure, carbn dixide exists as a gas. In ther wrds, neither slid nr liquid CO can exist at this temerature and ressure. Clearly, f the three hases, gaseus CO must be the mst stable and, hence, have the lwest free energy when T = 60 C and P ext =.00 atm. Gibbs Energy and Sntaneus Change 5. (E) Answer (b) is crrect. Br Br bnds are brken in this reactin, meaning that it is endthermic, with H 0. Since the number f mles f gas increases during the reactin, S 0. And, because G= H T S, this reactin is nn-sntaneus ( G 0 ) at lw temeratures where the H term redminates and sntaneus ( G 0 ) at high temeratures where the T S term redminates. 957

6. (E) Answer (d) is crrect. A reactin that rceeds nly thrugh electrlysis is a reactin that is nn-sntaneus. Such a reactin has G 0. 7. (E) (a) H 0 and S 0 (sincen gas 0 ) fr this reactin. Thus, this reactin is case f Table 9-. It is sntaneus at lw temeratures and nn-sntaneus at high temeratures. (b) We are unable t redict the sign f S fr this reactin, since n gas = 0. Thus, n strng redictin as t the temerature behavir f this reactin can be made. Since H > 0, we can, hwever, cnclude that the reactin will be nn-sntaneus at lw temeratures. (c) H 0 and S 0 (sincengas 0 ) fr this reactin. This is case f Table 9-. It is nn-sntaneus at lw temeratures, but sntaneus at high temeratures. 8. (E) (a) H 0 and S 0 (sincen gas 0 ) fr this reactin. This is case 4 f Table 9-. It is nn-sntaneus at all temeratures. (b) H 0 and S 0 (sincen gas 0 ) fr this reactin. This is case f Table 9-. It is sntaneus at all temeratures. (c) H 0 and S 0 (sincen gas 0 ) fr this reactin. This is case f Table 9-. It is sntaneus at lw temeratures and nn-sntaneus at high temeratures. 9. (E) First f all, the rcess is clearly sntaneus, and therefre G 0. In additin, the gases are mre disersed when they are at a lwer ressure and therefre S 0. We als cnclude that H = 0 because the gases are ideal and thus there are n frces f attractin r reulsin between them. 0. (E) Because an ideal slutin frms sntaneusly, G 0. Als, the mlecules f slvent and slute that are mixed tgether in the slutin are in a mre disersed state than the searated slvent and slute. Therefre, S 0. Hwever, in an ideal slutin, the attractive frces between slvent and slute mlecules equal thse frces between slvent mlecules and thse between slute mlecules. Thus, H = 0. There is n net energy f interactin.. (M) (a) An exthermic reactin (ne that gives ff heat) may nt ccur sntaneusly if, at the same time, the system becmes mre rdered (cncentrated) that is, S 0. This is articularly true at a high temerature, where the T S term dminates the G exressin. An examle f such a rcess is freezing water (clearly exthermic because the reverse rcess, melting ice, is endthermic), which is nt sntaneus at temeratures abve 0 C. (b) A reactin in which S 0 need nt be sntaneus if that rcess als is endthermic. This is articularly true at lw temeratures, where the H term dminates the G exressin. An examle is the varizatin f water (clearly an endthermic rcess, ne that requires heat, and ne that rduces a gas, s S 0), 958

which is nt sntaneus at lw temeratures, that is, belw00 C (assuming P ext =.00 atm).. (M) In this rblem we are asked t exlain whether the reactin AB(g)A(g)+B(g) is always ging t be sntaneus at high rather than lw temeratures. In rder t answer this questin, we need t determine the signs f H, S and cnsequently G. Recall that G H T S. Stewise arach: Determine the sign f S: We are generating tw mles f gas frm ne mle. The randmness f the system increases and S must be greater than zer. Determine the sign f H: In this reactin, we are breaking A-B bnd. Bnd breaking requires energy, s the reactin must be endthermic. Therefre, H is als greater than zer. Use G H T S t determine the sign f G : G H T S. Since H is sitive and S is sitive there will be a temerature at which T S will becme greater than H. The reactin will be favred at high temeratures and disfavred at lw temeratures. Cnversin athway arach: S fr the reactin is greater than zer because we are generating tw mles f gas frm ne mle. H fr the reactin is als greater than zer because we are breaking A-B (bnd breaking requires energy). Because G H T S, there will be a temerature at which T S will becme greater than H. The reactin will be favred at high temeratures and disfavred at lw temeratures. Standard Gibbs Energy Change. (M) H = H f NH 4 Cl s H NH f g H HCl g f 4.4kJ/ml (46. kj/ml 9. kj/ml) 76.0 kj/ml G G f NH 4 Cl s G NH f g G HCl g f 0.9kJ/ml (6.48 kj/ml 95.0 kj/ml) 9. kj/ml G H T S S H G T 76.0kJ/ml 9. kj/ml 98 000 J kj 4. (M) (a) G = Gf CH6g Gf CHg Gf Hg (b) 85 J ml =.8 kj/ml 09. kj/ml 0.00 kj/ml = 4.0 kj/ml G =G f SO g + G f O g G f SO g = 00. kj/ml+0.00 kj/ml 7. kj/ml= +4.8 kj/ml 959

(c) G =Gf Fes +4Gf HO ggf FeO4 s4gf Hg = 0.00 kj/ml + 4 8.6 kj/ml 05 kj/ml 4 0.00 kj/ml = 0 kj/ml + + (d) G =G f Al aq +Gf H ggf Al s6g f H aq 5. (M) (a) S =S POCl S PCl gs O g =.4 J/.7 J/05. J/ = 8.7 J/ G H T S (b) = 485 kj/ml + 0.00 kj/ml 0.00kJ/ml 6 0.00 kj/ml = 970. kj/ml = = 60.0 J 98 8.7 J/ = 5060 J = 506 kj The reactin rceeds sntaneusly in the frward directin when reactants and rducts are in their standard states, because the value f G is less than zer. + 6. (M) (a) S = S Br l +S HNO aq S H aq S Br aq S NO g = 5. J/ + 5.6 J/0 J/8.4 J/40. J/ =.6 J/ G = H TS = 6.60 J 98.6 J/ = +4.4 0 J = +4.4 kj (b) The reactin des nt rceed sntaneusly in the frward directin when reactants and rducts are in their standard states, because the value f G is greater than zer. 7. (M) We cmbine the reactins in the same way as fr any Hess's law calculatins. (a) N Og g g +08.4 kj= 04.kJ N N + O G = g g NO g + O g NO g +O Net: N Og G = +0.6kJ G = 04. +0.6 =.6 kj This reactin reaches an equilibrium cnditin, with significant amunts f all secies being resent. This cnclusin is based n the relatively small abslute value f G. (b) N +6H g g 4NH g G =.0 kj= 66.0 kj 4NH +5O g g 4NOg+6H Ol G = 00.5 kj 4NO g N +O g g G = +7. kj= 46. kj Net: 6H +O g g This reactin is three times the desired reactin, which therefre has G = 4.7 kj = 474. kj. The large negative G value indicates that this reactin will g t cmletin at 5 C. 6H Ol G = 66.0kJ 00.5kJ 46.kJ = 4.7kJ 960

(c) 4NH +5O g g 4NOg+6H Ol G = 00.5 kj 4NO g N +O g g G = +7. kj= 46. kj N g + O g N O g G = +08.4 kj G 4NH g + 4O g NO g + 6HO l = 00.5kJ 46. kj + 08.4 kj = 48.kJ This reactin is twice the desired reactin, which, therefre, has G = -574. kj. The very large negative value f the G fr this reactin indicates that it will g t cmletin. 8. (M) We cmbine the reactins in the same way as fr any Hess's law calculatins. (a) (b) COS+CO g g SO g+cog CO+H g Og CO g+h g COS+H g Og SO g+cog+h G = 46.4 kj= +46.6 kj G = 8.6 kj= 57. kj g G = +46.6 57. = +89.4 kj This reactin is sntaneus in the reverse directin, because f the large sitive value f G COS+CO g g SO g+cog G = 46.4 kj= +46.6 kj CO+H g Og CO g+h g G =8.6 kj= 85.8 kj COS+H g Og CO g+so g+h g G = +46.6 85.8 = +60.8 kj This reactin is sntaneus in the reverse directin, because f the large sitive value f G. (c) G G COS+H g Og CO g+h Sg COS g +H g CO g +H S g = +.4 kj CO g + H O g CO g + H g = 8.6 kj = 8.6 kj G =.4 kj 8.6 kj = 0.0 kj The negative value f the G fr this reactin indicates that it is sntaneus in the frward directin. 9. (D) The cmbustin reactin is :C 6 H 6 l+ 5 O g 6CO g+h Og r l (a) f f f 6 6 f 5 5 G =6G CO g + G H O l G CH l G O g = 6 94.4 kj + 7. kj +4.5 kj 0.00 kj = 0 kj 5 (b) G =6Gf COg +Gf HOg Gf CH 6 6l Gf Og 5 = 6 94.4 kj + 8.6 kj +4.5 kj 0.00 kj = 77 kj 96

We culd determine the difference between the tw values f G by nting the difference between the tw rducts: H Ol H Og and determining the value f G fr this difference: G = G f H Og Gf H Ol = 8.6 7. kj = 5.5 kj 0. (M) We wish t find the value f the H fr the given reactin: F g F(g) S = S F g S F g = 58.8 J 0.8 J = +4.8 J H = G + TS =.9 0 J + 98 4.8 J/ = 58. kj/mle f bnds The value in Table 0. is 59 kj/ml, which is in quite gd agreement with the value fund here.. (M) (a) S rxn = S rducts - S reactants = [ ml 0. J ml + ml 88.8 J ml ] [ ml 47.4 J ml + ml 8.5 J ml ] = 54.5 J S rxn = = 0.0545 kj (b) (c) H rxn = bnds brken in reactants (kj/ml)) bnds brken in rducts(kj/ ml)) = [4 ml (89 kj ml ) N-H + 4 ml ( kj ml ) O-F ] [4 ml (0 kj ml ) N-F + 4 ml (464 kj ml ) O-H ] H rxn = 66 kj G rxn = H rxn TS rxn = 66 kj 98 (0.0545 kj ) = 600 kj Since the G rxn is negative, the reactin is sntaneus, and hence feasible (at 5 C ). Because bth the entry and enthaly changes are negative, this reactin will be mre highly favred at lw temeratures (i.e., the reactin is enthaly driven). (D) In this rblem we are asked t find G at 98 fr the decmsitin f ammnium nitrate t yield dinitrgen xide gas and liquid water. Furthermre, we are asked t determine whether the decmsitin will be favred at temeratures abve r belw 98. In rder t answer these questins, we first need the balanced chemical equatin fr the rcess. Frm the data in Aendix D, we can determine H rxn and S rxn. Bth quantities will be required t determine G rxn (G rxn =H rxn -TS rxn ). Finally the magnitude f G rxn as a functin f temerature can be judged deending n the values f H rxn and S rxn. Stewise arach: First we need the balanced chemical equatin fr the rcess: NH 4 NO (s) N O(g) + H O(l) Nw we can determine H rxn by utilizing H f values rvided in Aendix D: H f NH 4 NO (s) N O(g) + H O(l) -65.6 kjml - 8.05 kjml - -85.6 kjml - H rxn =H f rducts H f reactants H rxn =[ ml(85.8 kj ml ) + ml(8.05 kj ml )] [ ml (65.6 kj ml )] 96

H rxn = 4.0 kj Similarly, S rxn can be calculated utilizing S values rvided in Aendix D NH 4 NO (s) N O(g) + H O(l) S 5. Jml - - 9.9 Jml - - 69.9 Jml - - S rxn =S rducts S reactants S rxn =[ ml 69.9 J ml + ml 9.9 J ml ] [ ml 5. J ml ] S rxn =08.6 J = 0.086 kj T find G rxn we can either utilize G f values rvided in Aendix D r G rxn =H rxn - TS rxn : G rxn =H rxn -TS rxn =-4.0 kj 98.5 0.086 kj - G rxn =-86. kj Magnitude f G rxn as a functin f temerature can be judged deending n the values f H rxn and S rxn : Since H rxn is negative and S rxn is sitive, the decmsitin f ammnium nitrate is sntaneus at all temeratures. Hwever, as the temerature increases, the TS term gets larger and as a result, the decmsitin reactin shift twards rducing mre rducts. Cnsequently, we can say that the reactin is mre highly favred abve 98 (it will als be faster at higher temeratures) Cnversin athway arach: Frm the balanced chemical equatin fr the rcess NH 4 NO (s) N O(g) + H O(l) we can determine H rxn and S rxn by utilizing H f and S values rvided in Aendix D: H rxn =[ ml(85.8 kj ml ) + ml(8.05 kj ml )] [ ml (65.6 kj ml )] H rxn = 4.0 kj S rxn =[ ml 69.9 J ml + ml 9.9 J ml ] [ ml 5. J ml ] S rxn =08.6 J = 0.086 kj G rxn =H rxn -TS rxn =-4.0 kj 98.5 0.086 kj - G rxn =-86. kj Since H rxn is negative and S rxn is sitive, the decmsitin f ammnium nitrate is sntaneus at all temeratures. Hwever, as the temerature increases, the TS term gets larger and as a result, the decmsitin reactin shift twards rducing mre rducts. The reactin is highly favred abve 98 (it will als be faster). The Thermdynamic Equilibrium Cnstant. (E) In all three cases, = because nly gases, ure slids, and ure liquids are resent eq in the chemical equatins. There are n factrs fr slids and liquids in eq exressins, and gases aear as artial ressures in atmsheres. That makes eq the same as fr these three reactins. 96

We nw recall that = c RT n. Hence, in these three cases we have: SO g + O g SO g ; n = + = ; = = RT (a) gas c (b) HI g H g + I g ; n = + = 0; = = gas c (c) 4 n =0 =+ = = RT 4. (M) (a) NH HCO s NH g + CO g + H O l ; gas c P{H g } 4 = P {H O g } 4 (b) Terms fr bth slids, Fe(s) and Fe O 4 s, are rerly excluded frm the thermdynamic equilibrium cnstant exressin. (Actually, each slid has an activity f.00.) Thus, the equilibrium artial ressures f bth H g and H Og d nt deend n the amunts f the tw slids resent, as lng as sme f each slid is resent. One way t understand this is that any chemical reactin ccurs n the surface f the slids, and thus is unaffected by the amunt resent. (c) We can rduce H g frm H Og withut regard t the rrtins f Fe(s) and Fe O 4 s with the qualificatin, that there must always be sme Fe(s) resent fr the rductin f H g t cntinue. 5. (M) In this rblem we are asked t determine the equilibrium cnstant and the change in Gibbs free energy fr the reactin between carbn mnxide and hydrgen t yield methanl. The equilibrium cncentratins f each reagent at 48 were rvided. We rceed by first determining the equilibrium cnstant. Gibbs free energy can be calculated using G RT ln. Stewise arach: First determine the equilibrium cnstant fr the reactin at 48: CO(g)+H (g) CHOH(g) [CH OH (g)] [CO(g)][H (g)] 0.0089 0.09 0.08 4.5 Nw use G RT ln t calculate the change in Gibbs free energy at 48 : G RT ln G 8.4 48 ln(4.5)jml -. 0 4 Jml - G kjml - Cnversin athway arach: [CH OH (g)] [CO(g)][H (g)] 0.0089 0.09 0.08 4.5 G RT ln 8.4 48 ln(4.5)jml -. 0 4 Jml - G kjml - 964

6. (M) Gibbs free energy fr the reactin ( G H T S ) can be calculated using H f and S values fr CO(g), H (g) and CH OH(g) frm Aendix D. H H f (CH OH(g)) [H f (CO(g)) H f (H (g)] H 00.7kJml - (0.5kJml - 0kJml - ) 90.kJml - S S (CH OH(g)) [S (CO(g)) S (H (g)] S 9.8J ml - (97.7J ml - 0.7J ml - ) 9.J - ml - G 90.kJml - 48 (9.)kJ- ml - 5.7kJml - 000 Equilibrium cnstant fr the reactin can be calculated using G RT ln ln G RT ln 5.7 000Jml 8.4J ml 48.9 e.9.0 0 The values are different because in this case, the calculated is the thermdynamic equilibrium cnstant that reresents the reactants and rducts in their standard states. In Exercise 5, the reactants and rducts were nt in their standard states. Relatinshis Invlving G, G, Q and 7. (M) G =Gf NO ggf NO g0.5 Gf O g = ln = 8.450 kj ml 98 ln = 86.55 kj/ml 04. kj/ml 0.5 0.00 kj/ml = 68.9 kj/ml RT 68.9 kj/ml ln = = 7.8 8.450 kj ml 98 8. (M) (a) G =Gf NO 5g Gf NO 4g Gf Og 7.8 =e =8 0 = 5. kj/ml 97.89 kj/ml 0.00 kj/ml = 4.4 kj/ml G 4.40 J/ml (b) G = RTln ln = = =.9 RT 8.45 J ml 98 =e =90.9 7 9. (M) We first balance each chemical equatin, then calculate the value f G with data frm Aendix D, and finally calculate the value f eq with the use f G = RTln. (a) 4HClg + Og HOg + Cls G =G H Og +G Cl g 4G HClg G O g f f f f kj kj kj kj kj = 8.6 + 0 4 95.0 0 = 76.0 ml ml ml ml ml G +76.00 J/ml ln = = = +0.7 = e = 0 RT 8.45 J ml 98 +0.7 965

(b) Fe O s +H g Fe O s +H Og 4 f 4 f f f G =G FeO s + G H O g G FeO s G H g = 05 kj/ml 8.6 kj/ml 74. kj/ml 0.00 kj/ml = kj/ml G 0 J/ml ln = = = ; = e = 4 0 RT 8.45 J ml 98 (c) + Ag aq +SO4 aq AgSO4s G = G f Ag SO 4 s G f Ag + aq G f SO 4 + 5 aq = 68.4 kj/ml 77.kJ/ml 744.5 kj/ml= 8.kJ/ml G 8.0 J/ml ln = = =.; = e = 80 RT 8.45 J ml 98 40. (E) S S S S S = {CO g }+ {H g } {CO g } {H O g } +. 4 =.7 J ml +0.7 J ml 97.7 J ml 88.8 J ml = 4. J ml 4. (M) In this rblem we need t determine in which directin the reactin SOg + Og SOg is sntaneus when the artial ressure f SO, O, and SO are.00-4, 0.0 and 0.0 atm, resectively. We rceed by first determining the standard free energy change fr the reactin ( G ) using tabulated data in Aendix D. Change in Gibbs free energy fr the reactin ( G ) is then calculated by emlying the equatin G G RTln Q,where Q is the reactin qutient. Based n the sign f G, we can determine in which directin is the reactin sntaneus. Stewise arach: First determine G fr the reactin using data in Aendix D: SOg + Og SOg G =G f SO g G SO f g G O f g G (7. kj/ml) (00. kj/ml) 0.0 kj/ml G 4.8kJ Calculate G by emlying the equatin G G RTln Q, where Q is the reactin qutient: G G RTln Q Q P{SO g } = P {O g } P {SO g } 966

Q 0.0 atm = 5.00 4 0.0 atm.00 atm G = 4.8 kj + (8.45 0 kj/ml )(98 )ln(5.0 0 6 ) G = 4.8 kj + 8. kj = 04 kj. Examine the sign f G t decide in which directin is the reactin sntaneus: Since G is negative, the reactin is sntaneus in the frward directin. Cnversin athway arach: G =G f SO g G SO f g G O f g G (7. kj/ml) (00. kj/ml) 0.0 kj/ml=-4.8kj G G RTln Q P{SO Q = } g P{O }P{SO g } g 0.0 atm 5.0 06 0.0 atm.0 0 4 atm G = 4.8 kj + (8.45 0 kj/ml )(98 )ln(5.0 0 6 )=-04 kj. Since G is negative, the reactin is sntaneus in the frward directin. 6 4. (M) We begin by calculating the standard free energy change fr the reactin: H g +Cl g HCl g G =Gf HCl g Gf Cl g Gf H g ( 95.0 kj/ml) 0.0 kj/ml 0.0 kj/ml 90.6kJ Nw we can calculate G by emlying the equatin G G RTln Q, where P{HClg } 0.5 atm Q = ; Q = P {H g } P {Cl g } 0.5 atm 0.5 atm G = 90.6 kj + (8.45 0 kj/ml )(98 )ln() G = 90.6 kj + 0 kj = 90.6 kj. Since G is negative, the reactin is sntaneus in the frward directin. 4. (M) In rder t determine the directin in which the reactin is sntaneus, we need t calculate the nn-standard free energy change fr the reactin. T accmlish this, we will emly the equatin G G RTln Q, where c Q c + [HO aq ] [CHCO aq ] = [CH CO H aq ] ; Q c.00 M =.0 0 (0.0 M) 5 G = 7.07 kj + (8.45 0 kj/ml )(98 )ln(.0 0 5 ) G = 7.07 kj + (8.5 kj) =.46 kj. Since G is negative, the reactin is sntaneus in the frward directin. 967

44. (M) As was the case fr exercise 9, we need t calculate the nn-standard free energy change fr the reactin. Once again, we will emly the equatin G G RTln Q, but this time + [NH4 aq ] [OH aq ].00 M 5 Qc = ; Qc =.0 0 [NH aq ] (0.0 M) G = 9.05 kj + (8.45 0 kj/ml)(98 )ln(.0 0 5 ) G = 9.05 kj + (8.5 kj) = 0.5 kj. Since G is sitive, the reactin is sntaneus in the reverse directin. 45. (E) The relatinshi S = (G H )/T (Equatin (b)) is incrrect. Rearranging this equatin t ut G n ne side by itself gives G = H + TS. This equatin is nt valid. The TS term shuld be subtracted frm the H term, nt added t it. 46. (E) The G value is a werful thermdynamic arameter because it can be used t determine the equilibrium cnstant fr the reactin at each and every chemically reasnable temerature via the equatin G = RT ln. 47. (M) (a) T determine we need the equilibrium artial ressures. In the ideal gas law, each artial ressure is defined by P = nrt / V. Because R, T, and V are the same fr each gas, and because there are the same number f artial ressure factrs in the numeratr as in the denminatr f the exressin, we can use the rati f amunts t determine. P CO(g) P H O(g) n CO(g) n H O(g) 0.4 ml CO 0.4 ml H O = = = = 0.659 P CO (g) P H (g) n CO (g) n H (g) 0.76 ml CO 0.76 ml H (b) G = RT ln = 8.45 J ml 000. ln 0.659 (c) Q 000 =.467 0 J/ml =.467 kj/ml 0.040 ml CO 0.0650ml H O = = 0. 0.659 = 0.0750 ml CO0.095 ml H Since Q is smaller than, the reactin will rceed t the right, frming rducts, t attain equilibrium, i.e., G = 0. n. Fr the reactin 48. (M) (a) We knw that = c RT SO g + O g SO g, = + =, and therefre a value f can be btained. n gas = c RT.8 0.4 0.0806 L atm 000 ml We recgnize that = since all f the substances invlved in the reactin are gases. We can nw evaluate G. 968

8.45 J G = RT ln 4 eq = 000 ln.4 =.00 J/ml = 0. kj/ml ml (b) We can evaluate Q c fr this situatin and cmare the value with that f c =.8 0 t determine the directin f the reactin t reach equilibrium. Q 0.7 ml SO [SO ].50 L c 45.80 [SO ] [O ] 0.40 ml SO 0.8 ml O.50 L.50 L Since Q c is smaller than c the reactin will shift right, rducing sulfur trixide and cnsuming sulfur dixide and mlecular xygen, until the tw values are equal. 49. (M) (a) = c G = RT ln = 8.450 kj ml 445 + 7 ln 50. =.4 kj eq n / = = c RT =.7 0 0.0898 = 8.4 0 G = RT ln = 8.450 kj ml 98 ln 8.40 g (b) G = +68.9 kj/ml n (c) + (d) = = c RT = 4.60 0.0806 98 = 0. G = RT ln = 8.45 0 kj ml 98 ln0.= +5.40 kj/ml 6 = c = 9.4 0 = ln = 8.45 0 kj ml G RT 98 ln 9.40 6 G = +8.7 kj/ml c 50. (M) (a) The first equatin invlves the frmatin f ne mle f Mg + aq + MgOH s and H aq, while the secnd equatin invlves the frmatin f nly half-a-mle f Mg + aq. We wuld exect a free energy change f half the size if nly half as much rduct is frmed. (b) c frm The value f fr the first reactin is the square f the value f fr the secnd reactin. The equilibrium cnstant exressins are related in the same fashin. = Mg+ = H + Mg + / H + = (c) The equilibrium slubilities will be the same regardless which exressin is used. The equilibrium cnditins (slubilities in this instance) are the same n matter hw we chse t exress them in an equilibrium cnstant exressin. 969

5. (E) G RT = ln = 8.450 kj ml 98 ln 6.50 = 67.4 kj/ml CO g + Cl g COCl g G = 67.4kJ/ml C grahite + O g CO g G f = 7. kj/ml C grahite + O g + Cl g COCl g G f = 04.6 kj/ml G f f COCl g given in Aendix D is 04.6 kj/ml, thus the agreement is excellent. 5. (M) In each case, we first determine the value f G fr the slubility reactin. Frm that, we calculate the value f the equilibrium cnstant,, fr the slubility reactin. + (a) AgBr s Ag aq + Br aq + G = Gf Ag aq + G f Br aq Gf AgBrs = 77. kj/ml 04.0 kj/ml 96.90 kj/ml = +70.0 kj/ml G 70.00 J/ml ln = = = 8.; s = e = 60 RT 8.45 J ml 98.5 (b) + CaSO4 s Ca aq +SO4 aq + G = G f Ca aq + G f SO4 aq Gf CaSO4s s 8. = 55.6 kj/ml 744.5 kj/ml kj/ml = +4 kj/ml G 40 J/ml ln = = = 4; s = e = 80 RT 8.45 J ml 98.5 (c) + Fe OH s Fe aq + OH aq G = G f Fe + aq +G f OH aq = 4.7 kj/ml + 57. kj/ml G f FeOH s 4 7 696.5 kj/ml= +0. kj/ml G 0. 0 J/ml ln = = = 88.8 s = e =.60 RT 8.45 J ml 98.5 88.8 9 5. (M)(a) We can determine the equilibrium artial ressure frm the value f the equilibrium cnstant. G 58.540 J/ml G = RTln ln = = =.6 RT 8.45 J ml 98.5 = P{O g } = e = 5.50 P{O g } = 5.50 =.0 0 atm /.6 (b) Lavisier did tw things t increase the quantity f xygen that he btained. First, he ran the reactin at a high temerature, which favrs the rducts (i.e., the side with mlecular xygen.) Secnd, the mlecular xygen was remved immediately after it was frmed, which causes the equilibrium t shift t the right cntinuusly (the shift twards rducts as result f the remval f the O is an examle f Le Châtelier's rincile). 970

54. (D) (a) We determine the values f H and S frm the data in Aendix D, and then the value f G at 5 C = 98. H = Hf CHOH g + Hf HO ghf CO ghf H g = 00.7 kj/ml + 4.8 kj/ml9.5 kj/ml 0.00 kj/ml = 49.0 kj/ml S = S CH OH g + S HO g S CO g S Hg = (9.8 +88.8.7 0.7) J ml = 77. J ml (b) = = 49.0 kj/ml 98 0.77 kj ml G H TS = +.8 kj/ml Because the value f G is sitive, this reactin des nt rceed in the frward directin at 5 C. Because the value f H is negative and that f S is negative, the reactin is nnsntaneus at high temeratures, if reactants and rducts are in their standard states. The reactin will rceed slightly in the frward directin, hwever, t rduce an equilibrium mixture with small quantities f CHOHg and H Og. Als, because the frward reactin is exthermic, this reactin is favred by lwering the temerature. That is, the value f increases with decreasing temerature. (c) G500 H T S = 9.60 J/ml = RT ln = = 49.0 kj/ml 500. 0.77 kj ml = 9.6 kj/ml G 9.60 J/ml ln 9.5; e 7.0 RT 8.45 J ml 500. (d) Reactin: CO + g H g 9.5 5 CH OHg +H Og Initial:.00 atm.00 atm 0 atm 0 atm Changes: x atm x atm +x atm +x atm Equil: (.00 x ) atm.00 x atm x atm x atm 5 PCHOH PHO x x =7.0 = = P CO P H.00 x.00 x x 5 7. 0 8.5 0 atm P CHOH x Our assumtin, that x.00 atm, is valid. G and as Functin f Temerature 55 (M)(a) S = S Na CO s + S H Ol + S CO g S NaHCO s J J J J J = 5.0 + 69.9 +.7 0.7 = 5. ml ml ml ml ml 97

(b) H = H Na CO s + H H Ol + H CO g H NaHCO s f f f f kj kj kj kj kj = 85.8 9.5 950.8 = +9 ml ml ml ml ml (c) = = 9 kj/ml 98 5. 0 kj ml G H TS = 9 kj/ml 64. kj/ml = 7 kj/ml (d) 7 0 J/ml 8.45 J ml 98 G G = RTln ln = = = 0.9 RT =e 0.9 = 0 5 56. (M) (a) S = S CH CH OH g + S H O g S CO g S H g S CH OH g S S J J J J J = 8.7 +88.8 97.7 0.7 9.8 ml ml ml ml ml J = 7.4 ml H = Hf CH CH OH g+ Hf HOg Hf COg Hf H ghf CH OH g kj kj kj kj kj H = 5. 4.8 0.5 0.00 00.7 ml ml ml ml ml kj H = 65.7 ml kj kj kj kj kj G = 65.7 98 7.4 0 = 65.4 + 67.8 = 97.9 ml ml ml ml ml (b) (c) H 0 fr this reactin. Thus it is favred at lw temeratures. Als, because n gas = + 4 =, which is less than zer, the reactin is favred at high ressures. First we assume that neither S nr H varies significantly with temerature. Then we cmute a value fr G at 750. Frm this value f G, we cmute a value fr. G = H T S = 65.7 kj/ml 750. 7.4 0 kj ml = 65.7 kj/ml + 70.6kJ/ml = +4.9kJ/ml = RT ln ln = G RT = 4. 9 0 J/ml 8.45 J ml 750. = 0.79 =e0.79 = 0.5 57. (E) In this rblem we are asked t determine the temerature fr the reactin between irn(iii) xide and carbn mnxide t yield irn and carbn dixide given G, H, and S. We rceed by rearranging G =H T S in rder t exress the temerature as a functin f G, H, and S. 97

Stewise arach: Rearrange G =H T S in rder t exress T as a functin f G, H, and S : G =H T S T S =H G T = H G S Calculate T: T = 4.8 0 J 45.5 0 J =.60 5. J/ Cnversin athway arach: G =H T S T H G 4.8 0 J 45.5 0 J S 5. J/ 5 58. (E) We use the van't Hff equatin with H =.8 0 J/ml, T = 800., T = 00. C= 7, and =9. 0. ln = H R T T =.8 05 J/ml 8.45 J ml 800 7 = = e =.9 0 = =.9 0 9.0 = 0 6 9.0 =.6 0 59. (M) We first determine the value f G at 400 C, frm the values f H and S, which are calculated frm infrmatin listed in Aendix D. H =Hf NH ghf N g Hf H g = 46.kJ/ml 0.00kJ/ml 0.00 kj/ml = 9.kJ/ml N S =S NH g S N g S H g = 9.5 J ml 9.6J ml 0.7 J ml = 98.7 J ml G = H T S = 9. kj/ml 67 0.987 kj ml = +4.5 kj/ml = RT ln G 4.50 J/ml ln = = = 7.4; = e = 6.00 RT 8.45J ml 67 7.4 4 97

60. (M) (a) H = Hf COg + Hf H ghf CO ghf HOg = 9.5 kj/ml 0.00 kj/ml 0.5 kj/ml 4.8 kj/ml = 4. kj/ml S = S CO g + S H g S CO g S HO g =.7 J ml +0.7 J ml 97.7 J ml 88.8 J ml = 4. J ml (b) G H T S 4.kJ/ml 98.5 (4. 0 )kj/ml G 4.kJ/ml.6kJ/ml 8.6kJ/ml G = H T S = 4. kj/ml 875 4. 0 kj ml = 4. kj/ml + 6.8 kj/ml = 4.4 kj/ml = RT ln G 4.40 J/ml RT 8.45J ml 875 +0.60 ln = = = +0.60 = e =.8 6. (M) We assume that bth H and S are cnstant with temerature. H =H f SO g H f SO g H f O g = 95.7 kj/ml 96.8 kj/ml 0.00 kj/ml= 97.8 kj/ml S =S SO g S SO g S O g = 56.8 J ml 48. J ml 05. J ml = 87.9 J ml G = H T S = RT ln H = T S RT ln T = 97.80 J/ml T = 650 6 87.9 J ml 8.45 J ml ln.00 H S Rln This value cmares very favrably with the value f T =6.7 0 that was btained in Examle 9-0. 6. (E) We use the van't Hff equatin t determine the value f H (448 C = 7 and 50 C = 6 ). ln = H R T T =ln50.0 H = 0.9 = 66.9 R 6 7 = H. 04 R H 0.9 = =.0 ; 4 - R. 0 H - - =.0 8.45 J ml - = 0 J ml - = kj ml 974

6. (M) (a) H 57.0 J/ml ln = = =. R T T 8.45 J ml 98 7. = e = 0. = 0. 0. = 0.04 at 7 (b) ln = H R T 98 T 98 T T = 57. 0 J/ml 8.45 J ml T 98 =ln0..00 =.80 =.80 8.45 57. 0 =.7 0 4 4 4 =.70 =.60.70 =.04 0 ; T = 9 64. (D) First we calculate G at 98 t btain a value fr eq at that temerature. f f f G =G NO g G NO g G O g = 5. kj/ml 86.55 kj/ml 0.00 kj/ml = 70.48 kj/ml G 70.480 J/ml 8.4 ln 8.4 e. 0 RT 8.45 J 98.5 ml Nw we calculate H fr the reactin, which then will be inserted int the van't Hff equatin. H =H f NO g H f NOg H f O g =.8 kj/ml 90.5 kj/ml 0.00 kj/ml = 4.4 kj/ml H 4.40 J/ml ln = = = 9.6 R T T 8.45 J ml 98 7 9.6 5 5 8 = e = 9.50 ; = 9.50.0 =. 0 Anther way t find at 00 ºC is t cmute H 4.4 kj/ml frm H f values S 46.5 J ml frm S values. Then determine G (59.5 kj/ml), and and find with the exressin G = RT ln. Nt surrisingly, we btain the same 8 result, =. 0. 65. (M) First, the van't Hff equatin is used t btain a value f H. 00 C = 47 and 60 C = 5. H.50 H ln = = ln = 6.56 = 8 R T T 4.560 8.45 J ml 5 47 5 6.56 5 6.56 =.9 0 H H = =.0 J/ml=.0 kj/ml 5.9 0 975

Anther rute t H is the cmbinatin f standard enthalies f frmatin. CO g + H g CH g + H O g 4 = f CH4g + f HOg f COg f Hg H H H H H = 74.8 kj/ml 4.8 kj/ml 0.5 0.00 kj/ml = 06. kj/ml Within the recisin f the data sulied, the results are in gd agreement. 66. (D) (a) t, C T, / T, ln 0. 0.0 0 50..0 0 70. 4.9 0 00. 7.68 0.66 0 5.006.90 0 4 7.849 6.7 0 5.07.0.465 Plt f ln( ) versus /T 0.006 0.009 /T(-) 0.00 0 - ln -4-6 -8-0 y = -540.x + 9.8 - The sle f this grah is H / R =.54 0 4 4 H = 8.45 J ml.540 = 8 0 J/ml = 8 kj/ml (b) When the ttal ressure is.00 atm, and bth gases have been rduced frm NaHCO s, P{H O g }= P {CO g }=.00 atm = P{H O g } P {CO g } =.00.00 =.00 Thus, ln =ln.00= 0.000. The crresnds t / T =.59 0 ; T = 86. We can cmute the same temerature frm the van't Hff equatin. 976

ln = H R T T = 8 0 J/ml 8.45 J ml T 0 =ln.66 05.00 =.006 T 0 =.006 8.45 8 0 = 7.5 0 4 T 4 4 = 7.5 0 =.0 0 7.5 0 =.59 0 ; T = 86 0 This temerature agrees well with the result btained frm the grah. Culed Reactins 67. (E) (a) We cmute G fr the given reactin in the fllwing manner H = Hf TiCl4 l + Hf O g Hf TiO s Hf Cl g = 804. kj/ml + 0.00 kj/ml 944.7 kj/ml 0.00 kj/ml = +40.5 kj/ml S = S TiCl4 l + S O g S TiO s S Cl g = 5. J ml + 05. J ml 50. J ml. J ml = 9. J ml G = H T S = +40.5 kj/ml 98 9.0 kj ml = +40.5 kj/ml +.6 kj/ml = +5. kj/ml (b) Thus the reactin is nn-sntaneus at 5 C. (we als culd have used values f G f t calculate G ). Fr the cited reactin, G =G f CO g G f COg G f G =94.4 kj/ml Then we cule the tw reactins. 7. kj/ml 0.00 kj/ml = 54.4 kj/ml TiO s + Cl g TiCl l + O g 4 CO g + O g CO g G 4 O g G = +5. kj/ml G = 54.4 kj/ml TiO s + Cl g + CO g TiCl l + CO g ; = 6. kj/ml The culed reactin has G 0, and, therefre, is sntaneus. 977

68. (E) If G 0 fr the sum f culed reactins, the reductin f the xide with carbn is sntaneus. (a) NiOs Nis + O g Cs+ Og COg Net : NiO s+ Cs Nis + COg Therefre the culed reactin is sntaneus (b) MnOs Mn s + O g Cs+ Og COg Mn s + COg Net : MnO s +C s Therefre the culed reactin is nn-sntaneus G = +5 kj G = 50 kj G = +5 kj 50 kj = 5 kj G = +80 kj G = 50 kj G = +80 kj 50 kj = +0 kj (c) TiOs Tis + Og G = +60 kj Cs + O g COg Net : TiO s + C s Ti s + CO g Therefre the culed reactin is nn-sntaneus G = 50 kj = 500 kj G = +60 kj 500 kj = +0 kj 69. (E) In this rblem we need t determine if the hshrylatin f arginine with ATP is a sntaneus reactin. We rceed by culing the tw given reactins in rder t calculate G t fr the verall reactin. The sign f G t can then be used t determine whether the reactin is sntaneus r nt. Stewise arach: First determine G t fr the culed reactin: ATP+H O ADP+P G t.5kjml - arginine+p hshrarginine+h O G t.kjml - ATP+arginine hshrarginine+adp G (.5.)kJml -.7kJml - Examine the sign f G t : G t 0. Therefre, the reactin is nt sntaneus. Cnversin athway arach: G t fr the culed reactin is: ATP+arginine hshrarginine+adp G (.5.)kJml -.7kJml - Since G t 0,the reactin is nt sntaneus. 978

70. (E) By culing the tw reactins, we btain: Glu - +NH 4 + Gln+H O G t 4.8kJml - ATP+H O ADP+P G t.5kjml - Glu - +NH 4 + +ATP Gln+ADP+P G (4.8.5)kJml - 6.7kJml - Therefre, the reactin is sntaneus. INTEGRATIVE AND ADVANCED EXERCISES 7. (M) (a) The nrmal biling int f mercury is that temerature at which the mercury var ressure is.00 atm, r where the equilibrium cnstant fr the varizatin equilibrium reactin has a numerical value f.00. This is als the temerature where G 0, since G RT ln eq and ln (.00) 0. Hg(l) Hg(g) H Hf [Hg(g)] Hf [Hg(l)] 6. kj/ml 0.00 kj/ml 6. kj/ml S S[Hg(g)] S[Hg(l)] 75.0J ml 76.0J ml 99.0J ml 0 H T S 6.0 6.0 T 99.0 J ml J/ml 69 J/ml T 99.0 J ml (b) The var ressure in atmsheres is the value f the equilibrium cnstant, which is related t the value f the free energy change fr frmatin f Hg var. Gf [ Hg(g)].8 kj/ml RT ln eq.8 0 J/ml.84 6 ln.84 e.65 0 atm 8.45 J ml 98.5 Therefre, the var ressure f Hg at 5ºC is.65 0-6 atm. 7. (M) (a) TRUE; It is the change in free energy fr a rcess in which reactants and rducts are all in their standard states (regardless f whatever states might be mentined in the statement f the rblem). When liquid and gaseus water are each at atm at 00 C (the nrmal biling int), they are in equilibrium, s that G = G = 0 is nly true when the difference f the standard free energies f rducts and reactants is zer. A reactin with G = 0 wuld be at equilibrium when rducts and reactants were all resent under standard state cnditins and the ressure f H O(g) =.0 atm is nt the standard ressure fr H O(g). (b) FALSE; G 0. The system is nt at equilibrium. 979

(c) FALSE; G can have nly ne value at any given temerature, and that is the value crresnding t all reactants and rducts in their standard states, s at the nrmal biling int G = 0 [as was als the case in answering art (a)]. Water will nt varize sntaneusly under standard cnditins t rduce water var with a ressure f atmsheres. (d) TRUE; G 0. The rcess f transfrming water t var at.0 atm ressure at 00 C is nt a sntaneus rcess; the cndensatin (reverse) rcess is sntaneus. (i.e. fr the system t reach equilibrium, sme H O(l) must frm) 7. (D) G G Br (g)] G [Cl (g)] G [BrCl(g)] f [ f f (0.00 kj/ml) ( 0.98 kj/ml) (.kj/ml).54 kj/ml RT ln G.540 J/ml ln RT 8.45J ml 98.5.0.0 e 0.6 Fr ease f slving the rblem, we duble the reactin, which squares the value f the equilibrium cnstant. (0.57) eq 0. 0 Reactin: BrCl(g) Br (g) Cl (g) Initial:.00 ml 0 ml 0 ml Changes: xml xml xml Equil: (.00 x)ml xml xml P{Br (g)} P{Cl (g)} [ n{br (g)} RT / V ][ n{cl (g)} RT / V ] n{br (g)} n{cl (g)} P{BrCl(g)} [ n{brcl(g)} RT / V ] n{brcl(g)} x x (0.6) 0.6 (.00 x).00 x x 0.6 0.7 x 0.6 x 0.0mlBr 0.0 mlcl.7.00 x 0.580 ml BrCl 74. (M) First we determine the value f fr the dissciatin reactin. If I (g) is 50% dissciated, then fr every mle f undissciated I (g), ne mle f I (g) has dissciated, rducing tw mles f I(g). Thus, the artial ressure f I(g) is twice the artial ressure f I (g) I(g) I(g) ). ( P.00 atm PI (g) PI(g) PI (g) PI (g) PI (g) PI (g) ttal PI(g) (0.667).4 ln 0.9 PI (g) 0. H H f [I(g)] H f S S[I(g)] S[I (g)] 80.8J ml 60.7 J ml Nw we equate tw exressins fr G and slve fr T. 0.atm [I (g)] 06.8 kj/ml 6.44 kj/ml 5. kj/ml 00.9J ml 980