SPONTANEITY, ENTROPY, AND FREE ENERGY

Transcription

1 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY Questins. Living rganisms need an external surce f energy t carry ut these prcesses. Green plants use the energy frm sunlight t prduce glucse frm carbn dixide and water by phtsynthesis. In the human bdy, the energy released frm the metablism f glucse helps drive the synthesis f prteins. Fr all prcesses cmbined, ΔS univ must be greater than zer (the secnd law).. Dispersin increases the entrpy f the universe because the mre widely smething is dispersed, the greater the disrder. We must d wrk t vercme this disrder. In terms f the secnd law, it wuld be mre advantageus t prevent cntaminatin f the envirnment rather than t clean it up later. As a substance disperses, we have a much larger area that must be decntaminated.. As a prcess ccurs, S univ will increase; S univ cannt decrease. ime, like S univ, nly ges in ne directin. 4. he intrductin f mistakes is an effect f entrpy. he purpse f redundant infrmatin is t prvide a cntrl t check the "crrectness" f the transmitted infrmatin. 5. his reactin is kinetically slw but thermdynamically favrable (ΔG < 0). hermdynamics nly tells us if a reactin can ccur. answer the questin will it ccur, ne als needs t cnsider the kinetics (speed f reactin). he ultravilet light prvides the activatin energy fr this slw reactin t ccur. 6. a. hermdynamics is cncerned with nly the initial and final states f a reactin, and nt the pathway required t get frm reactants t prducts. It is the kinetics f a reactin that cncentrates n the pathway and speed f a reactin. Fr these plts, the reactin with the smallest activatin energy will be fastest. he activatin energy is the energy reactants must have in rder t vercme the energy barrier t cnvert t prducts. S the fastest reactin is reactin 5 with the smallest activatin. Reactins,, and 4 all shuld have the same speed because they have the same activatin energy. (his assumes all ther kinetic factrs are the same.) Reactin will be the slwest since it has the largest activatin energy. b. If the prducts have lwer energy than the reactants, then the reactin is exthermic. Reactin,,, and 5 are exthermic. If the prducts have higher energy than the reactants, then the reactin is endthermic. Only reactin 4 is endthermic. Nte that it is the thermdynamics f a reactin that dictates whether a reactin is exthermic r endthermic. he kinetics plays n factr in these designatins. 659

2 660 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY c. he thermdynamics f reactin determine ptential energy change. In an exthermic reactin, sme f the ptential energy stred in chemical bnds is cnverted t thermal energy (heat is released as the ptential energy decreases). he exthermic reactin with the greatest lss f ptential energy is the reactin with the largest energy difference (E) between reactants and prducts. Reactins and 5 bth have the same largest E values, s reactins and 5 have the greatest change in ptential energy. Reactins and bth have the same smallest E values, s reactins and have the smallest change in ptential energy. 7. S surr = H/; heat flw (H) int r ut f the system dictates S surr. If heat flws int the surrundings, the randm mtins f the surrundings increase, and the entrpy f the surrundings increases. he ppsite is true when heat flws frm the surrundings int the system (an endthermic reactin). Althugh the driving frce described here really results frm the change in entrpy f the surrundings, it is ften described in terms f energy. Nature tends t seek the lwest pssible energy. 8. When slid NaCl disslves, the Na + and Cl ins are randmly dispersed in water. he ins have access t a larger vlume and a larger number f pssible psitins. Hence psitinal disrder has increased and S is psitive. Fr a salt disslving in water, H is usually a value clse t zer (see Sectin. f the text). Smetimes H is psitive and smetimes H is negative fr a salt disslving in water, but the magnitude f H is small. S the nly predictin that can be made is that H fr this prcess will be clse t zer. 9. Nte that these substances are nt in the slid state but are in the aqueus state; water mlecules are als present. here is an apparent increase in rdering when these ins are placed in water as cmpared t the separated state. he hydrating water mlecules must be in a highly rdered arrangement when surrunding these anins. 0. G = R ln K = H S ; HX(aq) H + (aq) + X (aq) K a reactin; the value f K a fr HF is less than ne, while the ther hydrgen halide acids have K a >. In terms f G, HF must have a psitive G rxn value, while the ther HX acids have G rxn < 0. he reasn fr the sign change in the K a value, between HF versus HCl, HBr, and HI is entrpy. S fr the dissciatin f HF is very large and negative. here is a high degree f rdering that ccurs as the water mlecules assciate (hydrgen bnd) with the small F ins. he entrpy f hydratin strngly ppses HF dissciating in water, s much s that it verwhelms the favrable hydratin energy making HF a weak acid.. One can determine S and H fr the reactin using the standard entrpies and standard enthalpies f frmatin in Appendix 4; then use the equatin G = H S. One can als use the standard free energies f frmatin in Appendix 4. And finally, ne can use Hess s law t calculate G. Here, reactins having knwn G values are manipulated t determine G fr a different reactin. Fr temperatures ther than 5 C, G is estimated using the G = H S equatin. he assumptins made are that the H and S values determined frm Appendix 4 data are temperature-independent. We use the same H and S values as determined when =

3 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 66 5 C; then we plug in the new temperature in Kelvin int the equatin t estimate G at the new temperature.. he sign f G tells us if a reactin is spntaneus r nt at whatever cncentratins are present (at cnstant and ). he magnitude f G equals w max. When G < 0, the magnitude tells us hw much wrk, in thery, culd be harnessed frm the reactin. When G > 0, the magnitude tells us the minimum amunt f wrk that must be supplied t make the reactin ccur. G gives us the same infrmatin nly when the cncentratins fr all reactants and prducts are at standard cnditins ( atm fr gases, M fr slute). hese cnditins rarely ccur. G = R ln K; frm this equatin, ne can calculate K fr a reactin if G is knwn at that temperature. herefre, G gives the equilibrium psitin fr a reactin. determine K at a temperature ther than 5 C, ne needs t knw G at that temperature. We assume H and S are temperature-independent and use the equatin G = H S t estimate G at the different temperature. Fr K =, we want G = 0, which ccurs when H = S. Again, assume H and S are temperature-independent; then slve fr (= H /S ). At this temperature, K = because G = 0. his nly wrks fr reactins where the signs f H and S are the same (either bth psitive r bth negative). When the signs are ppsite, K will always be greater than (when H is negative and S is psitive) r K will always be less than (when H is psitive and S is negative). When the signs f H and S are ppsite, K can never equal.. he light surce fr the first reactin is necessary fr kinetic reasns. he first reactin is just t slw t ccur unless a light surce is available. he kinetics f a reactin are independent f the thermdynamics f a reactin. Even thugh the first reactin is mre favrable thermdynamically (assuming standard cnditins), it is unfavrable fr kinetic reasns. he secnd reactin has a negative ΔG value and is a fast reactin, s the secnd reactin which ccurs very quickly is favred bth kinetically and thermdynamically. When cnsidering if a reactin will ccur, thermdynamics and kinetics must bth be cnsidered. 4. Using Le Chatelier's principle, a decrease in pressure (vlume increases) will favr the side with the greater number f particles. hus I(g) will be favred at lw pressure. Exercises Lking at ΔG: ΔG = ΔG + R ln ( I / I ); ln( I / I ) 0 when I I = 0 atm and ΔG is psitive (nt spntaneus). But at I = I = 0.0 atm, the lgarithm term is negative. If R ln Q > ΔG, then ΔG becmes negative, and the reactin is spntaneus. Spntaneity, Entrpy, and the Secnd Law f hermdynamics: Free Energy 5. a, b, and c; frm ur wn experiences, salt water, clred water, and rust frm withut any utside interventin. It takes an utside energy surce t clean a bedrm, s this prcess is nt spntaneus. 6. c and d; it takes an utside energy surce t build a huse and t launch and keep a satellite in rbit, s these prcesses are nt spntaneus.

4 66 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 7. ssible arrangements fr ne mlecule: Bth are equally prbable. ssible arrangements fr tw mlecules: way way way ways, way mst prbable ssible arrangement fr three mlecules: way ways ways way equally mst prbable 8. a. he mst likely arrangement wuld be mlecules in each flask. here are 0 different ways (micrstates) t achieve gas mlecules in each flask. Let the letters A-F represent the six mlecules. he different ways t have grups f three are: ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF, BCD, BCE, BCF, BDE, BDF, BEF, CDE, CDF, CEF, and DEF. b. With six mlecules and tw bulbs, there are 6 = 64 pssible arrangements. Because there are 0 ways t achieve gas mlecules in each bulb, the prbability expressed as a percent f this arrangement is: 0 00 =.5% We draw all the pssible arrangements f the tw particles in the three levels. kj x x xx kj x xx x 0 kj xx x x tal E = 0 kj kj kj kj kj 4 kj he mst likely ttal energy is kj.

5 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY kj AB B A B A kj AB B A A B 0 kj AB A B A B E ttal = 0 kj kj 4 kj kj kj kj kj kj kj he mst likely ttal energy is kj.. a. H at 00 C and 0.5 atm; higher temperature and lwer pressure means greater vlume and hence larger psitinal prbability. b. N ; N at S has the greater vlume because is smaller and is larger. c. H O(l) has a larger psitinal prbability than H O(s).. Of the three phases (slid, liquid, and gas), slids are mst rdered (have the smallest psitinal prbability) and gases are mst disrdered (have the largest psitinal prbability). hus a, b, and f (melting, sublimatin, and biling) invlve an increase in the entrpy f the system since ging frm a slid t a liquid r frm a slid t a gas r frm a liquid t a gas increases disrder (increases psitinal prbability). Fr freezing (prcess c), a substance ges frm the mre disrdered liquid state t the mre rdered slid state; hence, entrpy decreases. rcess d (mixing) invlves an increase in disrder (an increase in psitinal prbability), while separatin (phase e) increases rder (decreases psitinal prbability). S, f all the prcesses, a, b, d, and f result in an increase in the entrpy f the system.. a. Biling a liquid requires heat. Hence this is an endthermic prcess. All endthermic prcesses decrease the entrpy f the surrundings (ΔS surr is negative). b. his is an exthermic prcess. Heat is released when gas mlecules slw dwn enugh t frm the slid. In exthermic prcesses, the entrpy f the surrundings increases (ΔS surr is psitive). 4. a. ΔS surr = ( kj) = 7.45 kj/k = J/K 98K b. ΔS surr = kj = 0.76 kj/k = 76 J/K 98K 5. ΔG = ΔS; when ΔG is negative, then the prcess will be spntaneus. a. ΔG = ΔS = 5 0 J (00. K)(5.0 J/K) = 4,000 J; nt spntaneus b. ΔG = 5,000 J (00. K)(00. J/K) = 5000 J; spntaneus c. Withut calculating ΔG, we knw this reactin will be spntaneus at all temperatures. is negative and ΔS is psitive (ΔS < 0). ΔG will always be less than zer with these sign cmbinatins fr and ΔS. d. ΔG = J (00. K)( 40. J/K) = 000 J; spntaneus

6 664 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 6. ΔG = ΔS; a prcess is spntaneus when ΔG < 0. Fr the fllwing, assume and ΔS are temperature-independent. a. When and ΔS are bth negative, ΔG will be negative belw a certain temperature where the favrable term dminates. When ΔG = 0, then = ΔS. Slving fr this temperature: = ΔS 8,000J =.0 0 K 60. J / K At <.0 0 K, this prcess will be spntaneus (ΔG < 0). b. When and ΔS are bth psitive, ΔG will be negative abve a certain temperature where the favrable ΔS term dminates. = 8,000J =.0 0 K ΔS 60. J / K At >.0 0 K, this prcess will be spntaneus (ΔG < 0). c. When is psitive and ΔS is negative, this prcess can never be spntaneus at any temperature because ΔG can never be negative. d. When is negative and ΔS is psitive, this prcess is spntaneus at all temperatures because ΔG will always be negative. 7. At the biling pint, ΔG = 0, s = ΔS. ΔS = 7.5 kj/ ml = 8.9 (7 5) K 8. At the biling pint, ΔG = 0, s = ΔS. = 0 kj/k ml = 89. J/K ml ΔS J/ml = 69.7 K 9.9J/K ml 9. a. NH (s) NH (l); ΔG = ΔS = 5650 J/ml 00. K (8.9 J/K ml) ΔG = 5650 J/ml 5780 J/ml = 0 J/ml Yes, NH will melt because ΔG < 0 at this temperature. b. At the melting pint, ΔG = 0, s = ΔS 5650 J/ml = 96 K. 8.9J/K ml 40. C H 5 OH(l) C H 5 OH(g); at the biling pint, G = 0 and S univ = 0. Fr the vaprizatin prcess, S is a psitive value, whereas H is a negative value. calculate S sys, we will determine S surr frm H and the temperature; then S sys = S surr fr a system at equilibrium.

7 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 665 S surr = H 8.70 J/ml = 0. J/K ml 5K S sys = S surr = (0.) = 0. J/K ml Chemical Reactins: Entrpy Changes and Free Energy 4. a. Decrease in psitinal prbability; ΔS will be negative. here is nly ne way t achieve gas mlecules all in ne bulb, but there are many mre ways t achieve the gas mlecules equally distributed in each flask. b. Decrease in psitinal prbability; ΔS is negative fr the liquid t slid phase change. c. Decrease in psitinal prbability; ΔS is negative because the mles f gas decreased when ging frm reactants t prducts ( mles 0 mles). Changes in the mles f gas present as reactants are cnverted t prducts dictates predicting psitinal prbability. he gaseus phase always has the larger psitinal prbability assciated with it. d. Increase in psitinal prbability; ΔS is psitive fr the liquid t gas phase change. he gas phase always has the larger psitinal prbability. 4. a. Decrease in psitinal prbability (Δn < 0); ΔS () b. Decrease in psitinal prbability (Δn < 0); ΔS () c. Increase in psitinal prbability; ΔS (+) d. Increase in psitinal prbability; ΔS (+) 4. a. C graphite (s); diamnd has a mre rdered structure (has a smaller psitinal prbability) than graphite. b. C H 5 OH(g); the gaseus state is mre disrdered (has a larger psitinal prbability) than the liquid state. c. CO (g); the gaseus state is mre disrdered (has a larger psitinal prbability) than the slid state. 44. a. He (0 K); S = 0 at 0 K b. N O; mre cmplicated mlecule, s has the larger psitinal prbability. c. NH (l); the liquid state is mre disrdered (has a larger psitinal prbability) than the slid state. 45. a. H S(g) + SO (g) S rhmbic (s) + H O(g); because there are mre mlecules f reactant gases than prduct mlecules f gas (Δn = < 0), ΔS will be negative.

8 666 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY ΔS = n psprducts n S r reactants ΔS = [ ml S rhmbic (s) ( J/K ml) + ml H O(g) (89 J/K ml)] ΔS = 474 J/K 660. J/K = 86 J/K [ ml H S(g) (06 J/K ml) + ml SO (g) (48 J/K ml)] b. SO (g) SO (g) + O (g); because Δn f gases is psitive (Δn = ), ΔS will be psitive. ΔS = ml(48 J/K ml) + ml(05 J/K ml) [ ml(57 J/K ml)] = 87 J/K c. Fe O (s) + H (g) Fe(s) + H O(g); because Δn f gases = 0 (Δn = ), we can t easily predict if ΔS will be psitive r negative. ΔS = ml(7 J/K ml) + ml(89 J/K ml) [ ml(90. J/K ml) + ml (4 J/K ml)] = 8 J/K 46. a. H (g) + / O (g) H O(l); since Δn f gases is negative, ΔS will be negative. ΔS = ml H O(l) (70. J/K ml) ΔS = 70. J/K 4 J/K = 64 J/K [ ml H (g) ( J/K ml) + / ml O (g) (05 J/K ml)] b. CH OH(g) + O (g) CO (g) + 4 H O(g); because Δn f gases is psitive, ΔS will be psitive. [ ml (4 J/K ml) + 4 ml (89 J/K ml)] [ ml (40. J/K ml) + ml (05 J/K ml)] = 89 J/K c. HCl(g) H + (aq) + Cl (aq); the gaseus state dminates predictins f ΔS. Here, the gaseus state is mre disrdered than the ins in slutin, s ΔS will be negative. ΔS = ml H + (0) + ml Cl (57 J/K ml) ml HCl(87 J/K ml) = 0. J/K CF4 H CH F 47. C H (g) + 4 F (g) CF 4 (g) + H (g); ΔS = S S [S 4S ] CF 4 58 J/K = ( ml) S + J/K [0 J/K + 4(0 J/K)], S CF 4 = 6 J/K ml CO SO O CS 48. CS (g) + O (g) CO (g) + SO (g); ΔS = S S [S S ] CS 4 J/K = 4 J/K + (48 J/K) (05 J/K) ( ml) S, S = 8 J/K ml CS

9 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY a. S rhmbic S mnclinic ; this phase transitin is spntaneus (ΔG < 0) at temperatures abve 95 C. ΔG = ΔS; fr ΔG t be negative nly abve a certain temperature, then is psitive and ΔS is psitive (see able 7.5 f text). b. Because ΔS is psitive, S rhmbic is the mre rdered crystalline structure (has the smaller psitinal prbability) (s,α) 4 (s,β) a. At < 76.9 C, this reactin is spntaneus, and the sign f ΔG is (). At 76.9 C, ΔG = 0, and abve 76.9 C, the sign f ΔG is (+). his is cnsistent with () and ΔS (). b. Because the sign f ΔS is negative, the β frm has the mre rdered structure (has the smaller psitinal prbability). 5. a. When a bnd is frmed, energy is released, s is negative. here are mre reactant mlecules f gas than prduct mlecules f gas (Δn < 0), s ΔS will be negative. b. ΔG = ΔS; fr this reactin t be spntaneus (ΔG < 0), the favrable enthalpy term must dminate. he reactin will be spntaneus at lw temperatures (at a temperature belw sme number), where the term dminates. 5. Because there are mre prduct gas mlecules than reactant gas mlecules (Δn > 0), ΔS will be psitive. Frm the signs f and ΔS, this reactin is spntaneus at all temperatures. It will cst mney t heat the reactin mixture. Because there is n thermdynamic reasn t d this, the purpse f the elevated temperature must be t increase the rate f the reactin, that is, kinetic reasns. 5. a. CH 4 (g) + O (g) CO (g) + H O(g) f 75 kj/ml ΔG f 5 kj/ml Data frm Appendix 4 S 86 J/K ml p f, prducts r f, reactants ΔS = n psprducts = n n ; n S r reactants = ml(4 kj/ml) + ml(9.5 kj/ml) [ ml(75 kj/ml)] = 80 kj ΔS = ml(89 J/K ml) + ml(4 J/K ml) [ ml(86 J/K ml) + ml(05 J/K ml)] = 4 J/K here are tw ways t get ΔG. We can use ΔG = ΔS (be careful f units): ΔG = ΔS = 80 0 J (98 K)( 4 J/K) = J = 80 kj r we can use ΔG f values, where ΔG = n ΔG n ΔG p f, prducts r f, reactants :

10 668 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY ΔG = ml(9 kj/ml) + ml(94 kj/ml) [ ml(5 kj/ml)] ΔG = 80 kj (Answers are the same within rund ff errr.) b. 6 CO (g) + 6 H O(l) C 6 H O 6 (s) + 6 O (g) f 9.5 kj/ml S 4 J/K ml = 75 [686) + 6(9.5)] = 80 kj ΔS = 6(05) + [6(4) + 6(70.)] = 6 J/K ΔG = 80 kj (98 K)( 0.6 kj/k) = 880. kj c. 4 O 0 (s) + 6 H O(l) 4 H O 4 (s) f (kj/ml) S (J/K ml) = 4 ml(79 kj/ml) [ ml(984 kj/ml) + 6 ml(86 kj/ml)] = 46 kj ΔS = 4(0.) [9 + 6(70.)] = 09 J/K ΔG = ΔS = 46 kj (98 K)( 0.09 kj/k) = 54 kj d. HCl(g) + NH (g) NH 4 Cl(s) f (kj/ml) S (J/K ml) = 4 [9 46] = 76 kj; ΔS = 96 [87 + 9] = 84 J/K ΔG = ΔS = 76 kj (98 K)( 0.84 kj/k) = 9 kj 54. a. = (46 kj) = 9 kj; ΔS = (9 J/K) [( J/K) + 9 J/K] = 99 J/K ΔG = ΔS = 9 kj 98 K(0.99 kj/k) = kj b. ΔG is negative, s this reactin is spntaneus at standard cnditins.

11 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 669 c. ΔG = 0 when = ΔS 9kJ = 460 K 0.99kJ/ K At < 460 K and standard pressures ( atm), the favrable term dminates, and the reactin is spntaneus (ΔG < 0). 55. ΔG = 58.0 kj (98 K)( kj/k) = 5.40 kj ΔG = 0 = ΔS, = ΔS 58.0 kj = 8.6 K 0.766kJ/ K ΔG is negative belw 8.6 K, where the favrable term dminates. 56. H O(l) H O(g); ΔG = 0 at the biling pint f water at atm and 00. C. = ΔS, ΔS = J / ml = 09 J/K ml 7K At 90. C: ΔG = ΔS = 40.6 kj/ml (6 K)(0.09 kj/k ml) =.0 kj/ml As expected, ΔG > 0 at temperatures belw the biling pint f water at atm (prcess is nnspntaneus). At 0. C: ΔG = ΔS = 40.6 kj/ml (8 K)(0.09 J/K ml) =. kj/ml When ΔG < 0, the biling f water is spntaneus at atm, and > 00. C (as expected). 57. CH 4 (g) H (g) + C(s) ΔG = (5 kj) H (g) + O (g) H O(l) ΔG = 474 kj) C(s) + O (g) CO (g) ΔG = 94 kj CH 4 (g) + O (g) H O(l) + CO (g) ΔG = 87 kj C(s) + 6 O (g) 6 CO (g) ΔG = 6(94 kj) H (g) + / O (g) H O(l) ΔG = (7 kj) 6 CO (g) + H O(l) C 6 H 6 (l) + 5/ O (g) ΔG = / (699 kj) 6 C(s) + H (g) C 6 H 6 (l) ΔG = 5 kj 59. ΔG = n pδgf, prducts nrδg f, reactants, 74 kj = 05 kj Δ G f, SF4 Δ G f, SF4 = 7 kj/ml kj = 8(94 kj) + 0(7 kj), C H f, C H ΔG, ΔG = 6 kj/ml f a. ΔG = ml(0) + ml(9 kj/ml) [ ml(740. kj/ml) + ml(0)] = 5 kj

12 670 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY b. Because ΔG is psitive, this reactin is nt spntaneus at standard cnditins and 98 K. c. ΔG = ΔS, ΔS = H G 00. kj 5kJ = 0.6 kj/k 98K We need t slve fr the temperature when ΔG = 0: ΔG = 0 = ΔS, = ΔS, = H S 00.kJ 0.6kJ/K = 60 K his reactin will be spntaneus (ΔG < 0) at > 60 K, where the favrable entrpy term will dminate. 6. a. ΔG = (70. kj) (50 kj) = 464 kj b. Because ΔG is psitive, this reactin is nt spntaneus at standard cnditins at 98 K. c. ΔG = ΔS, = ΔG + ΔS = 464 kj + 98 K(0.79 kj/k) = 57 kj We need t slve fr the temperature when ΔG = 0: ΔG = 0 = ΔS, = 57 kj = 890 K ΔS 0.79kJ/ K his reactin will be spntaneus at standard cnditins (ΔG < 0) when > 890 K. Here the favrable entrpy term will dminate. 6. CH 4 (g) + CO (g) CH CO H(l) = 484 [75 + (9.5)] = 6 kj; ΔS = 60. (86 + 4) = 40. J/K ΔG = ΔS = 6 kj (98 K)(0.40 kj/k) = 56 kj At standard cncentratins, this reactin is spntaneus nly at temperatures belw = /ΔS = 67 K (where the favrable term will dminate, giving a negative ΔG value). his is nt practical. Substances will be in cndensed phases and rates will be very slw at this extremely lw temperature. CH OH(g) + CO(g) CH CO H(l) = 484 [0.5 + (0)] = 7 kj; ΔS = 60. ( ) = 78 J/K ΔG = 7 kj (98 K)(0.78 kj/k) = 90. kj his reactin als has a favrable enthalpy and an unfavrable entrpy term. But this reactin, at standard cncentratins, is spntaneus at temperatures belw = /ΔS = 6 K (a much higher temperature than the first reactin). S the reactin f CH OH and CO will be preferred at standard cncentratins. It is spntaneus at high enugh temperatures that the rates f reactin shuld be reasnable.

13 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY C H 4 (g) + H O(g) CH CH OH(l) = 78 (5 4) = 88 kj; ΔS = 6 (9 + 89) = 47 J/K When ΔG = 0, = ΔS, s = ΔS 88 0 J = 60 K. 47J/K Since the signs f and ΔS are bth negative, this reactin at standard cncentratins will be spntaneus at temperatures belw 60 K (where the favrable term will dminate). C H 6 (g) + H O(g) CH CH OH(l) + H (g) = 78 (84.7 4) = 49 kj; ΔS = + 6 ( ) = 7 J/K Bth and ΔS have unfavrable signs. ΔG can never be negative when is psitive and ΔS is negative. S this reactin can never be spntaneus at standard cnditins. hus the reactin C H 4 (g) + H O(g) C H 5 OH(l) wuld be preferred at standard cnditins. Free Energy: ressure Dependence and Equilibrium 65. ΔG = ΔG + R ln Q; fr this reactin: ΔG = ΔG + R ln NO NO ΔG = ml(5 kj/ml) + ml(0) [ ml(87 kj/ml) + ml(6 kj/ml)] = 98 kj O O ΔG = 98 kj J/K ml (98K) ln 000J/kJ (.00 0 ( )(.00 0 ) ( ) ) ΔG = 98 kj kj = 88 kj 66. ΔG = (0) + (9) [(4) + (00.)] = 90. kj ΔG = ΔG + R ln H S H O SO (8.45)(98) (0.00) = 90. kj + kj ln (.0 0 )(0.00) ΔG = 90. kj kj = 50. kj N 67. ΔG = ΔG + R ln Q = ΔG + R ln NO O4 ΔG = ml(98 kj/ml) ml(5 kj/ml) = 6 kj a. hese are standard cnditins, s ΔG = ΔG because Q = and ln Q = 0. Because ΔG is negative, the frward reactin is spntaneus. he reactin shifts right t reach equilibrium.

14 67 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY b. ΔG = 6 0 J J/K ml (98 K) ln ΔG = 6 0 J J = (0.) Because ΔG = 0, this reactin is at equilibrium (n shift). c. ΔG = J J/K ml (98 K) ln (0.9) ΔG = 6 0 J J =. 0 J = 0 J Because ΔG is psitive, the reverse reactin is spntaneus, and the reactin shifts t the left t reach equilibrium. NH 68. a. ΔG = ΔG + R ln N H ; ΔG =, NH ΔG f = (7) = 4 kj (8.45J/K ml)(98k) (50.) ΔG = 4 kj + ln 000J/kJ (00.)(00.) ΔG =4 kj kj = 67 kj (8.45J/K ml)(98k) (00.) b. ΔG = 4 kj + ln 000J/kJ (00.)(600.) ΔG = 4 kj 4.4 kj = 68 kj 69. NO(g) + O (g) NO (g) + O (g); ΔG = p f, prducts Σn ΔG Σn ΔG r f, reactants ΔG = ml(5 kj/ml) [ ml(87 kj/ml) + ml(6 kj/ml)] = 98 kj ΔG = R ln K, K ΔG exp R 5 (.98 0 J) exp 8.45J/K ml(98k) e 79.9 = Nte: When determining expnents, we will rund ff after the calculatin is cmplete. his helps eliminate excessive rund ff errr. 70. ΔG = ml(9 kj/ml) [ ml(4 kj/ml) + ml(00. kj/ml)] = 90. kj K ΔG exp R 4 ( J) exp 8.45J/K ml(98k) e 6. = ΔG = ΔS ; because there is a decrease in the number f mles f gaseus particles, ΔS is negative. Because ΔG is negative, must be negative. he reactin will be spntaneus at lw temperatures (the favrable term dminates at lw temperatures).

15 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY At 5.0 C: ΔG = ΔS = J/ml (98. K)( 76.6 J/K ml) = J/ml ΔG ΔG = R ln K, ln K = R K = e.66 = 8.7 ( J/ml) exp =.66 (8.45J/K ml)(98.k) At 00.0 C: ΔG = J/ml (7. K)( 76.6 J/K ml) = J/ml ln K = ( J/ml) =.540, K = e = (8.45J/K ml)(7.k) Nte: When determining expnents, we will rund ff after the calculatin is cmplete. his helps eliminate excessive rund ff errr. 7. a. ΔG = (9.) 78. = kj; = (4.).8 = 59. kj ΔS H G 59.kJ kj = 0. kj/k = J/K 98K ΔG 495,400J b. ΔG = R ln K, ln K R 8.45J/K ml(98k) K = e = c. Assuming and ΔS are temperature-independent: G 000 = 59. kj 000. K(0. kj/k) = 7 kj ln K = ( 7,000J) 8.45J/K ml(000. K) = 4.94, K = e 4.94 = When reactins are added tgether, the equilibrium cnstants are multiplied tgether t determine the K value fr the final reactin. H (g) +O (g) H O (g) K =. 0 6 H O(g) H (g) + /O (g) K = ( ) / H O(g) + /O (g) H O (g) K =. 0 6 ( ) / = ΔG = R ln K = 8.45J K ml (600. K) ln(5.4 0 ) = J/ml = 40 kj/ml

16 674 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 74. a. f (kj/ml) S (J/K ml) NH (g) 46 9 O (g) 0 05 NO(g) 90. H O(g) 4 89 NO (g) HNO (l) H O(l) NH (g) + 5 O (g) 4 NO(g) + 6 H O(g) = 6( 4) + 4(90.) [4( 46)] = 908 kj ΔS = 4() + 6(89) [4(9) + 5(05)] = 8 J/K ΔG = 908 kj (98 K)(0.8 kj/k) = 96 kj ΔG = R ln K, ln K = ΔG R ( 96 0 J) = J/K ml 98K ln K =.0 lg K, lg K = 68, K = 0 68 (an extremely large number) NO(g) + O (g) NO (g) = (4) [(90.)] = kj; ΔS = (40.) [() + (05)] = 47 J/K ΔG = kj (98 K)( 0.47 kj/k) = 68 kj K = exp ΔG ( 68,000J) exp R 8.45J/K ml(98k) = e 7.44 = 8. 0 Nte: When determining expnents, we will rund ff after the calculatin is cmplete. NO (g) + H O(l) HNO (l) + NO(g) = ( 74) + (90.) [(4) + ( 86)] = 74 kj ΔS = (56) + () [(40.) + (70.)] = 67 J/K ΔG = 74 kj (98 K)( 0.67 kj/k) = 6 kj K = exp ΔG 6000J exp R 8.45J/K ml(98k) =.4 e = 9 0 b. ΔG = R ln K; = 85 C = (85 + 7) K = 098 K; we must determine ΔG at 098 K. Δ G 098 = ΔS = 908 kj (098 K)(0.8 kj/k) = 07 kj

17 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 675 ΔG K = exp R 6 (.07 0 J) exp 8.45J/K ml (098K) = e.58 = c. here is n thermdynamic reasn fr the elevated temperature because is negative and ΔS is psitive. hus the purpse fr the high temperature must be t increase the rate f the reactin. ( 0,500J) 75. a. ΔG = R ln K, K = exp = J/K ml 98K b. C 6 H O 6 (s) + 6 O (g) 6 CO (g) + 6 H O(l) ΔG = 6 ml(94 kj/ml) + 6 ml(7 kj/ml) ml(9 kj/ml) = 875 kj 875kJ mlglucse mla = 94. ml A; 94. mlecules A/mlecule glucse 0.5kJ his is an verstatement. he assumptin that all the free energy ges int this reactin is false. Actually, nly 8 mles f A are prduced by metablism f mle f glucse. 76. a. ln K = ΔG R 4,000J = 5.65, K = e 5.65 =.5 0 (8.45J/K ml)(98k) b. Glutamic acid + NH Glutamine + H O ΔG = 4 kj A + H O AD + H O 4 ΔG = 0.5 kj Glutamic acid + A + NH Glutamine + AD + H O 4 ΔG = = 7 kj ln K = ΔG R ( 7,000J) = 6.86, K = e 6.86 = J/K ml(98k) NF (0.48) 77. K = 0.0(0.06) N F = Δ G 800 = R ln K = 8.45 J/K ml (800. K) ln ( ) = J/ml = 7 kj/ml 78. SO (g) + O (g) SO (g); ΔG = ( 7 kj) [( 00. kj)] = 4 kj ΔG = R ln K, ln K = K = e 57. = ΔG R ( 4.000J) = J/K ml (98K) K = SO (.0) =, (0.50) SO O SO SO =.0 0 atm

18 676 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY Frm the negative value f ΔG, this reactin is spntaneus at standard cnditins. here are mre mlecules f reactant gases than prduct gases, s ΔS will be negative (unfavrable). herefre, this reactin must be exthermic ( < 0). When and ΔS are bth negative, the reactin will be spntaneus at relatively lw temperatures where the favrable term dminates. ΔS 79. he equatin ln K = is in the frm f a straight line equatin R R (y = mx + b). A graph f ln K versus / will yield a straight line with slpe = m = /R and a y intercept = b = ΔS /R. Frm the plt: slpe = Δy Δx K 0 =. 0 4 K. 0 4 K = /R, =. 0 4 K 8.45 J/K ml =. 0 5 J/ml y intercept = 40. = ΔS /R, ΔS = J/K ml = 0 J/K ml As seen here, when is psitive, the slpe f the ln K versus / plt is negative. When is negative as in an exthermic prcess, then the slpe f the ln K versus / plt will be psitive (slpe = /R). 80. he ln K versus / plt gives a straight line with slpe = H /R and y intercept = S /R K = H /R, H = (8.45 J/K ml) ( K) H = J/ml =.4 kj/ml 4.5 = S /R, S = (4.5)(8.45 J/K ml) = 0.6 J/K ml Nte that the signs fr and ΔS make sense. When a bnd frms, < 0 and ΔS < 0. Additinal Exercises 8. Frm Appendix 4, S = 98 J/K ml fr CO(g) and S = 7 J/K ml fr Fe(s). Let S l = S fr Fe(CO) 5 (l) and S g = S fr Fe(CO) 5 (g). S = 677 J/K = ml(s l ) [ ml (7 J/K ml) + 5 ml(98 J/ K ml] S l = 40. J/K ml S = 07 J/K = ml (S g ) ml (40. J/K ml) S g = S fr Fe(CO) 5 (g) = 447 J/K ml

19 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY When an inic slid disslves, ne wuld expect the disrder f the system t increase, s ΔS sys is psitive. Because temperature increased as the slid disslved, this is an exthermic prcess, and ΔS surr is psitive (ΔS surr = /). Because the slid did disslve, the disslving prcess is spntaneus, s ΔS univ is psitive. 8. ΔS will be negative because ml f gaseus reactants frm ml f gaseus prduct. Fr ΔG t be negative, must be negative (exthermic). Fr exthermic reactins, K decreases as increases, resulting in a smaller quantity f prduct and a larger quantity f reactants present at equilibrium. herefre, the rati f the partial pressure f Cl 5 (a prduct) t the partial pressure f Cl (a reactant) will decrease when is raised. vap 84. At biling pint, ΔG = 0 s ΔS = ; fr methane: ΔS = J/ml Fr hexane: ΔS = = 84.5 J/ml K. V met = 4K J / ml K = 7. J/ml K nr V met =.00 ml(0.0806l atm/k ml)(k) = 9.9 L; V hex =.00atm nr = R(4 K) = 8. L Hexane has the larger mlar vlume at the biling pint, s hexane shuld have the larger entrpy. As the vlume f a gas increases, psitinal disrder increases. 85. slid I slid II; equilibrium ccurs when ΔG = 0. ΔG = ΔS, = ΔS, = /ΔS = 74.J/ml 7.0J/K ml = 4.7 K = 9.5 C 86. a. ΔG = R ln K = (8.45 J/K ml)(98 K) ln = J/ml = 6.0 kj/ml b. H O H + Cl O Cl H O Cl On each side f the reactin there are H O bnds and O Cl bnds. Bth sides have the same number and type f bnds. hus 0. c. ΔG = ΔS, ΔS = ΔG J = 0. J/K 98K d. Fr H O(g), Δ H f = 4 kj/ml and S = 89 J/K ml. = 0 =, HOCl f [ ml(4 kj/ml) + ml(80. J/K/ml)], Δ H f, HOCl = 8 kj/ml 0. J/K = S HOCl HOCl [ ml(89 J/K ml) + ml(66. J/K ml)], S = 8 J/K ml e. Assuming and ΔS are -independent: Δ G 500 = 0 (500. K)( 0. J/K) = J

20 678 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY ΔG = R ln K, K = G exp R exp = (8.45)(500.).4 e = HO HOCl f. ΔG = ΔG + R ln ; frm part a, ΔG = 6.0 kj/ml. ClO We shuld express all partial pressures in atm. Hwever, we perfrm the pressure cnversin the same number f times in the numeratr and denminatr, s the factrs f 760 trr/atm will all cancel. hus we can use the pressures in units f trr. ΔG = 6.0 kj/ml (8.45J/K 000J/kJ ml)(98k) (0.0) ln (8)(.0) = = 4 kj/ml 87. Ba(NO ) (s) Ba + (aq) + NO (aq) K = K sp ; ΔG = 56 + (09) (797) = 8 kj ΔG = R ln K sp, ln K sp = K sp = 7.6 e = Δ G 8,000J R 8.45J/K ml(98 K) = K + (bld) K + [K (muscle) ΔG = 0; ΔG = R ln [K ΔG = 8.45J K ml ] m ] b ; ΔG = w max (0. K) ln 0.5, ΔG = J/ml = 8.8 kj/ml At least 8.8 kj f wrk must be applied t transprt ml K +. Other ins will have t be transprted in rder t maintain electrneutrality. Either anins must be transprted int the cells, r catins (Na + ) in the cell must be transprted t the bld. he latter is what happens: [Na + ] in bld is greater than [Na + ] in cells as a result f this pumping. 89. HgbO Hgb + O ΔG = (70 kj) Hgb + CO HgbCO ΔG = 80 kj HgbO + CO HgbCO + O ΔG = 0 kj ΔG ΔG = R ln K, K = exp R ( 0 0 J) exp = 60 (8.45J/K ml)(98k) 90. HF(aq) H + (aq) + F [H ][F ] (aq); ΔG = ΔG + R ln [HF] 4 ΔG = R ln K = (8.45 J/K ml)(98 K) ln(7. 0 ) = J/ml

21 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 679 a. he cncentratins are all at standard cnditins, s ΔG = ΔG = J/ml (Q =.0 and ln Q = 0). Because ΔG is psitive, the reactin shifts left t reach equilibrium. b. ΔG = (.7 0 ) J/ml + (8.45 J/K ml)(98 K) ln 0.98 ΔG = J/ml J/ml = 0 ΔG = 0, s the reactin is at equilibrium (n shift). c. ΔG = (.0 0 ) J/ml (98 K) ln d. ΔG = (0.7) (98) ln =. 0 4 J/ml; shifts right = = 0; at equilibrium e. ΔG = (0.67) (98) ln = 0 J/ml; shifts left ΔS is mre favrable (less negative) fr reactin than fr reactin, resulting in K > K. In reactin, seven particles in slutin are frming ne particle in slutin. In reactin, fur particles are frming ne, which results in a smaller decrease in psitinal prbability than fr reactin. 9. A graph f ln K versus / will yield a straight line with slpe equal t /R and y intercept equal t ΔS /R. emp ( C) (K) 000/ (K ) K w ln K w

22 680 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY he straight-line equatin (frm a calculatr) is ln K = Slpe = K = R = (6.9 0 K 8.45 J/K ml) = J/ml y intercept = 9.09 = ΔS, ΔS = J/K ml = 75.6 J/K ml R 9. ΔG = R ln K; when K =.00, ΔG = 0 since ln.00 = 0. ΔG = 0 = ΔS = (4 kj) [86 kj] = 00. kj; ΔS = [(7 J/K) + (89 J/K)] = ΔS, = H S 00.kJ 0.8kJ/K = 75 K [90. J/K + ( J/K)] = 8 J/K 94. Equilibrium ccurs when the minimum in free energy has been reached. Fr this reactin, the minimum in free energy is when / f A has reacted. Because equilibrium lies clser t reactants fr this reactin, the equilibrium cnstant will be less than (K < ) which dictates that ΔG is greater than 0 (ΔG > 0; ΔG = RlnK). A(g) B(g) K p = Initial.0 atm 0 Change x +x Equil..0 x x Frm the plt, equilibrium ccurs when / f the A(g) has reacted. x = amunt A reacted t reach equilibrium = /(.0 atm) =.0 atm, x = 0.50 atm B A x 0.50 K p 0. 5= 0. (t sig. figs.) (.0 x) (.0) ChemWrk rblems he answers t the prblems 95-0 (r a variatin t these prblems) are fund in OWL. hese prblems are als assignable in OWL. Challenge rblems 0 a. Vessel : At 0C, this system is at equilibrium, s ΔS univ = 0 and ΔS = ΔS surr. Because the vessel is perfectly insulated, q = 0, s ΔS surr = 0 = ΔS sys.

23 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 68 b. Vessel : he presence f salt in water lwers the freezing pint f water t a temperature belw 0C. In vessel, the cnversin f ice int water will be spntaneus at 0C, s ΔS univ > 0. Because the vessel is perfectly insulated, ΔS surr = 0. herefre, ΔS sys must be psitive (ΔS > 0) in rder fr ΔS univ t be psitive. 04. he liquid water will evaprate at first and eventually equilibrium will be reached (physical equilibrium). Because evapratin is an endthermic prcess, is psitive. Because H O(g) is mre disrdered (greater psitinal prbability), ΔS is psitive. he water will becme cler (the higher energy water mlecules leave), thus Δ water will be negative. he vessel is insulated (q = 0), s ΔS surr = 0. Because the prcess ccurs, it is spntaneus, s ΔS univ is psitive. 05. O (g) O (g); = (4 kj) = 86 kj; ΔG = (6 kj) = 6 kj ln K = Δ G 6 0 J R (8.45J/K ml)(98 K) =.57, K =.57 e = We need the value f K at 0. K. Frm Sectin 7.8 f the text: ln K = ΔG R ΔS R Fr tw sets f K and : ΔS ln K = ; R R ln K = R Subtracting the first expressin frm the secnd: ln K ln K = H R r K ln K H R ΔS R 58 Let K = 7. 0, = 98 K; K = K 0, = 0. K; = 86 0 J 7. 0 ln K = K 0 58 = e 4. = , K 0 =. 0 7 K 0 =. 7 0 = O O O, O =. (.0 0 atm) 4 0 atm he vlume ccupied by ne mlecule f zne is:

24 68 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY V = nr = (/6.0 0 ml)( L atm/k (. 0 4 atm) ml)(0. K) = L Equilibrium is prbably nt maintained under these cnditins. When nly tw zne mlecules are in a vlume f L, the reactin is nt at equilibrium. Under these cnditins, Q > K, and the reactin shifts left. But with nly zne mlecules in this huge vlume, it is extremely unlikely that they will cllide with each ther. At these cnditins, the cncentratin f zne is nt large enugh t maintain equilibrium. 06. Arrangement I and V: S = k ln W; W = ; S = k ln = 0 Arrangement II and IV: W = 4; S = k ln 4 =.8 Arrangement III: W = 6; S = k ln 6 =.47 0 J/K ln 4, S =.9 0 J/K 0 J/K 07. a. Frm the plt, the activatin energy f the reverse reactin is E a + (ΔG ) = E a ΔG (ΔG is a negative number as drawn in the diagram). k f = A exp E a R (E and a G kr = A exp R ), k k f r Ea A exp R (Ea G A exp R ) If the A factrs are equal: k k f r E exp R a (E a G R ) ΔG = exp R ΔG Frm ΔG = R ln K, K = exp ; because K and R same expressin, K = k f /k r. k k f r are bth equal t the b. A catalyst will lwer the activatin energy fr bth the frward and reverse reactins (but nt change ΔG ). herefre, a catalyst must increase the rate f bth the frward and reverse reactins. 08. At equilibrium: H nr V.0 0 mlecules L atm (98K) mlecules/ml K ml.00l 0 H = atm 0 he pressure f H decreased frm.00 atm t atm. Essentially all f the H and Br has reacted. herefre, HBr =.00 atm because there is a : mle rati between HBr and H in the balanced equatin. Because we began with equal mles f H and Br, we will 0 have equal mles f H and Br at equilibrium. herefre, = atm. H Br

25 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 68 HBr (.00) K = 0 ( ) H Br = ; assumptins gd. ΔG = R ln K = (8.45 J/K ml)(98 K) ln( ) = J/ml ΔS = H G 0,800J/ml ( K 5 J/ml) = 0 J/K ml 09. a. ΔG = B A G G =, = 7 J ΔG K = exp 7J = exp = 0. R (8.45J/K ml)(98k) b. When Q =.00 > K, the reactin shifts left. Let x = atm f B(g), which reacts t reach equilibrium. A(g) B(g) Initial.00 atm.00 atm Equil x.00 x K = B A x x = 0.,.00x = 0. + (0.)x, x = 0.50 atm B = = 0.50 atm; A = =.50 atm c. ΔG = ΔG + R ln Q = ΔG + R ln( B / A ) ΔG = 7 J + (8.45)(98) ln (0.50/.50) = 7 J 7 J = 0 (carrying extra sig. figs.) ΔS 0. Frm Exercise 79, ln K =. Fr K at tw temperatures and, the R R K equatin can be manipulated t give (see Exercise 05): ln K R.5 0 ln J/K ml 98K 48K 5.6 = ( ml/j)( ), = J/ml Fr K = 8.84 at = 5 C: ln 8.84 = ( J/ml) (8.45J/K ml)(98k) 4 ΔS 8.45J/K, ml ΔS = ΔS = 0 J/K ml We get the same value fr ΔS using K =.5 0 at = 48 K data.

26 684 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY ΔG = R ln K; when K =.00, then ΔG = 0 since ln.00 = 0. Assuming and ΔS d nt depend n temperature: ΔG = 0 = ΔS, = ΔS, = ΔS J/K J/ml = 0 K ml. K = CO ; t ensure Ag CO frm decmpsing, CO shuld be greater than K. Frm Exercise 79, ln K = R ΔS R. Fr tw cnditins f K and, the equatin is: K ln K R Let = 5 C = 98 K, K = 6. K ln 6. 0 K ln trr 8.45J/K 7., K trr; = 0. C = 8 K, K =? J/ml ml 98K 8K = e 7. =. 0, K = 7.5 trr prevent decmpsitin f Ag CO, the partial pressure f CO shuld be greater than 7.5 trr. L L. Frm the prblem, χ C H χ 6 6 CCl = We need the pure vapr pressures ( ) in rder t 4 calculate the vapr pressure f the slutin. Use the thermdynamic data t determine the pure vapr pressure values. C 6 H 6 (l) C 6 H 6 (g) K = C H C 6H6 at 5 C 6 6 ΔG rxn ΔGf, C H (g) ΔG 6 6 f, C6H6 (l) ΔG = R ln K, ln K = ΔG R = = 9.66 kj/ml 4.50 kj/ml = 5.6 kj/ml J/ml =.08 (8.45J/K ml)(98k) K = H = C e = 0.5 atm Fr CCl 4 : ΔG rxn ΔGf, CCl (g) ΔG 4 f, CCl4 (l) = kj/ml (65. kj/ml) = 4.6 kj/ml K = CCl 4 ΔG = exp 460J/ml = exp R = 0.55 atm 8.45J/K ml 98K L C H χ C H C6H6 = 0.500(0.5 atm) = atm; CCl 4 = 0.500(0.55 atm) = atm

27 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 685 V H χ = C 6 6 C H 6 6 = tt 0.065atm = 0.065atm atm = V χ CCl 4 = = NaCl(s) Na + (aq) + Cl (aq) K = K sp = [Na + ][Cl ] ΔG = [(6 kj) + ( kj)] (84 kj) = 9 kj = 9000 J ( 9000J) ΔG = R ln K sp, K sp = exp 8.45J/K ml 98K = 8 = 40 NaCl(s) Na + (aq) + Cl (aq) K sp = 40 Initial s = slubility (ml/l) 0 0 Equil. s s K sp = 40 = s(s), s = (40) / = 6. = 6 M = [Cl ] 4. ΔS surr = / = q / q = heat lss by ht water = mles mlar heat capacity Δ q =.00 0 mlho 75.4 J g H O (98. 6.) = J 8.0g K ml ΔS surr = ( K 5 J) = 9 J/K 5. HX(aq) H + (aq) + X (aq) K a = Initial 0.0 M ~0 0 Equil. 0.0 x x x [H ][X ] [HX] Frm prblem, x = [H + ] = = (.5 0 ) ; K a = =. 0 ΔG = R ln K = 8.45 J/K ml(98 K) ln(. 0 ) = J/ml = 6 kj/ml Integrative rblems 6. Use the thermdynamic data t calculate the biling pint f the slvent. At biling pint: ΔG = 0 = ΔS, =.90 0 J/ml = 5. K ΔS J/K ml

28 686 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY Δ = K b m, (55.4 K 5. K) =.5 K kg/ml(m), m = Mass slvent = 50. ml 0.879g ml kg 000g = 0. kg..5 = 0.84 ml/kg 0.84mlslute 4g Mass slute = 0. kg slvent = 5.7 g = 6 g slute kgslvent ml 7. Because the partial pressure f C(g) decreased, the net change that ccurs fr this reactin t reach equilibrium is fr prducts t cnvert t reactants. A(g) + B(g) C(g) Initial 0.00 atm 0.00 atm 0.00 atm Change +x +x x Equil x x 0.00 x Frm the prblem, C = atm = 0.00 x, x = atm he equilibrium partial pressures are: A = x = = 0.60 atm, B = (0.60) = 0. atm, and C = atm K = (0.) = 5. G = R ln K = 8.45 J/K ml(98 K) ln(5.) = 4. 0 J/ml = 4. kj/ml 8. G = H S = J 98 K(75 J/K) = 4,00 J G = R ln K, ln K = K = e = ΔG R 4,000J = J/K ml 98K B + H O BH + + OH K = K b = 5.7 Initial 0.5 M 0 ~0 Change x +x +x Equil. 0.5 x x x 0 5 K b = = [BH ][OH [B] ] = x 0.5 x x, x = [OH ] = M ph = lg(.68 0 ) =.57; poh = =.48; assumptins gd

29 CHAER 7 SONANEIY, ENROY, AND FREE ENERGY 687 Marathn rblem 9. a. ΔS will be negative because there is a decrease in the number f mles f gas. b. Because ΔS is negative, must be negative fr the reactin t be spntaneus at sme temperatures. herefre, ΔS surr is psitive. c. Ni(s) + 4 CO(g) Ni(CO) 4 (g) = 607 [4(0.5)] = 65 kj; ΔS = 47 [4(98) + (0.)] = 405 J/K d. ΔG = 0 = ΔS, = e. = 50. C + 7 = K ΔS 650 J = 407 K r 4 C 405J / K ΔG = 65 kj ( K)( kj/k) = 4 kj ln K = ΔG R ( 4,000J) =.66, K = e.66 = J/K ml(k) f. = 7 C + 7 = 500. K ΔG 500 = 65 kj (500. K)( kj/k) = 8 kj ln K = 8,000J = 9.4, K = (8. 45 J/K ml)(500.k) 9.4 e =. 0 4 g. he temperature change causes the value f the equilibrium cnstant t change frm a large value favring frmatin f Ni(CO) 4 t a small value favring the decmpsitin f Ni(CO) 4 int pure Ni and CO. his is exactly what is wanted in rder t purify a nickel sample. h. Ni(CO) 4 (l) Ni(CO) 4 (g) K = Ni(CO)4 At 4 C (the biling pint): ΔG = 0 = ΔS ΔS = H K J = 9. J/K At 5 C: ΔG 5 = ΔS = J 45 K(9. J/K) = 0,00 J ΔG = R ln K, ln K = ( 0,00J) 8.45 J/K ml(45k) =.858, K p = e.858 = 7.4 A maximum pressure f 7.4 atm can be attained befre Ni(CO) 4 (g) will liquify.