CHAPTER 22 THE ELEMENTS IN NATURE AND INDUSTRY

Transcription

1 CHAPTER THE ELEMENTS IN NATURE AND INDUSTRY.1 It is abundant in the universe since its simple atms were the first created after the Big Bang. H has the lwest density f any material, s it is nly weakly held by the gravitatinal field f the Earth.. Irn frms irn(iii) xide, Fe O (cmmnly knwn as hematite). Calcium frms calcium carbnate, CaCO, (cmmnly knwn as limestne). Sdium is cmmnly fund in sdium chlride (halite), NaCl. Zinc is cmmnly fund in zinc sulfide (sphalerite), ZnS.. a) Differentiatin refers t the prcesses invlved in the frmatin f the earth int regins (cre, mantle, and crust) f differing cmpsitin. Substances separated accrding t their densities, with the mre dense material in the cre and less dense in the crust. b) The fur mst abundant elements are xygen, silicn, aluminum, and irn in rder f decreasing abundance. c) Oxygen is the mst abundant element in the crust and mantle but is nt fund in the cre. Silicn, aluminum, calcium, sdium, ptassium, and magnesium are als present in the crust and mantle but nt in the cre..4 Grups IA and IIA and the left half f the transitin metals are usually fund as xides, while the right half f the transitin metals and mst f the p-blck metals are fund as sulfides. The mre electrnegative metals tend t frm sulfides, the less electrnegative nes, xides..5 a) Bauxite (impure Al(OH) ) b) Air (N ) c) Seawater (NaCl) d) Limestne (CaCO ) e) Seawater (NaCl).6 Aluminsilicates are thermdynamically very stable. Great amunts f energy are needed fr the prductin f a metal frm a silicate, making the prcess ecnmically unfavrable..7 Plants prduced O, slwly increasing the xygen cncentratin in the atmsphere and creating an xidative envirnment fr metals. Evidence f Fe(II) depsits pre-dating Fe(III) depsits suggests this hypthesis is true. The decay f plant material and its incrpratin int the crust increased the cncentratin f carbn in the crust and created large fssil fuel depsits..8 Carbn atms frm a large number f bnds, which allws fr the frmatin f a wide variety f cmplex mlecules necessary fr life. Since C atms are small, they frm strng bnds with ne anther, giving rganic mlecules stability, particularly with respect t reactin with O and H O..9 Fixatin refers t the prcess f cnverting a substance in the atmsphere int a frm mre readily usable by rganisms. Examples are the fixatin f carbn by plants in the frm f carbn dixide and f nitrgen by bacteria in the frm f nitrgen gas. Fixatin f carbn dixide gas by plants cnverts the CO int carbhydrates during phtsynthesis. Fixatin f nitrgen gas by nitrgen-fixing bacteria invlves the cnversin f N t ammnia and ammnium ins..10 If t much CO enters the atmsphere frm human activities (primarily frest clearing, decmpsitin f limestne, and burning f C-cntaining fuels), the pssibility exists fr an unnatural warming f the earth due t trapping f heat by the extra CO..11 The labeled categry in the figure is Human activity and cmbustin. Tw ther categries are 1: Prductin f cmpunds f Mg and Ca frm dlmite, and : Prductin f steel and aluminum. -1

2 .1 Atmspheric nitrgen is utilized by three fixatin pathways: atmspheric, industrial, and bilgical. Atmspheric fixatin requires high-temperature reactins (e.g., lightning) t cnvert N int NO and ther xidized species. Industrial fixatin invlves mainly the frmatin f ammnia, NH, frm N and H. Bilgical fixatin ccurs in nitrgen-fixing bacteria that live in the rts f legumes. Human activity is referred t as industrial fixatin. Human activity is a significant factr, cntributing abut 17% f the nitrgen remved..1 Because N-N single bnds are weak, cyclic and chain cmpunds cntaining these bnds are nt very stable..14 a) N gaseus phsphrus cmpunds are invlved in the phsphrus cycle, s the atmsphere is nt included in the phsphrus cycle. b) Tw rles f rganisms in phsphrus cycle: 1) Plants excrete acid frm their rts t cnvert PO 4 - ins int mre sluble H PO 4 - ins, which the plant can absrb. ) Thrugh excretin and decay after death, rganisms return sluble phsphate cmpunds t the cycle..15 a) Atmspheric fixatin requires energy frm lightning r fires; industrial fixatin requires high temperature t make the prcess prceed at a reasnable rate. b) ΔH fr NO and NO f are bth endthermic, s their frmatin requires energy. In cntrast, ΔH f fr NH is exthermic. The reactin is slw, hwever, and requires energy t increase its rate. c) The energy cmes frm a widespread lw-grade energy surce: sunlight. d) There wuld be less N in the atmsphere and prbably a greater bimass n the earth as a result f a lw activatin energy fr nitrgen fixatin..16 a) N (g) + H O(l) NO(g) + 4 H + (aq) + 4 e xidatin b) H O(l) + N O(g) NO (g) + 6 H + (aq) + 6 e xidatin c) NH (aq) + H O(l) NO (aq) + 7 H + (aq) + 6 e xidatin d) NO (aq) + H + (aq) + e NO (aq) + H O(l) reductin e) N (g) + 6 H O(l) NO (aq) + 1 H + (aq) + 10 e xidatin.17 a) First determine the amunt f F present in 100. kg f flurapatite, using the mlar mass f Ca 5 (PO 4 ) F (504.1 g/ml); take 15% f this amunt. Cnvert ml F t ml SiF 4, and use the ideal gas law t determine vlume f SiF 4 gas. 10 g Ca 5(PO 4) F 1 ml Ca 5(PO 4) F 1 ml F 15% Mles F = ( 100. kg Ca 5(PO 4) F) 1 kg Ca 5(PO 4) F g Ca 5(PO 4) F 1 ml Ca 5(PO 4) F 100% = ml F (unrunded) 1mlSiF4 Mles SiF 4 = ( ml F) 4mlF = ml SiF 4 (unrunded) V = nrt / P = = = 1.1 x 10 L L atm ml SiF K ml K ( ) ( ) ( + ) ( 1.00 atm) b) Assume that all f the flurine in sdium hexaflursilicate is available as F when redisslved in drinking water. First, calculate the mles f Na SiF 6 that frm frm 7.46 ml f SiF 4 accrding t the reactin stichimetry. Then, calculate the mass f F available frm the Na SiF 6 that is prduced in the reactin. 1 ml NaSiF 6 Mle Na SiF 6 = ( ml SiF4 ) ml SiF = ml Na SiF 6 4 Mass F = ( ml Na SiF ) ml F g F = = 4. x 10 g F 6-1 ml Na SiF6 1 ml F -

3 The definitin f ppm ( parts per millin ) states that 1 ppm F = (1 g F - / (10 6 g H O). If 1 g F will fluridate 10 6 g H O t a level f 1 ppm, hw many grams (cnverted t ml using density, then cnverted frm ml t L t m ) f water can 44 g F fluridate t the 1 ppm level? Necessary cnversin factrs: 1 m = 1000 L, density f water = 1.00 g/ml g HO ml HO 1cm 10 m Vlume = ( g F ) 1gF 1.00 g HO 1 ml 1 cm = = 4. x 10 m.18 a) The irn ins frm an insluble salt, Fe (PO 4 ), that decreases the yield f phsphrus. This salt is f limited value. b) Each mle f Ca (PO 4 ) cntains mles f P atms and prduces, at 100% yield, 0.5 ml f P 4. Use cnversin factr 1 metric tn (T) = 1 x 10 kg. 98% 10 kg 10 g 1mlCa (PO 4) ( 50. T) x 100% 1 T 1 kg g Ca (PO 4) Mass Ca (PO 4 ) = 1mlP4 1.88gP4 1kg 1T 90.% ml Ca (PO 4) 1 ml P 4 10 g 10 kg 100% = = 8.8 metric tns P 4.19 a) An re is a naturally ccurring mixture f gangue and mineral frm which an element can be prfitably extracted. b) A mineral is a naturally ccurring, hmgeneus, crystalline slid, with a well-defined cmpsitin. c) Gangue is the debris, such as sand rck and clay, assciated with an re. d) Brine is a cncentrated salt slutin used as a surce f Na, Cl, etc..0 a) Rasting invlves heating the mineral in air (O ) at high temperatures t cnvert the mineral t the xide. b) Smelting is the reductin f the metal xide t the free metal using heat and a reducing agent such as cke. c) Fltatin is a separatin prcess in which the re is remved frm the gangue by expliting the different abilities f the tw t interact with detergent. The gangue sinks t the bttm and the lighter re-detergent mix is skimmed ff the tp. d) Refining is the final step in the purificatin prcess t yield the pure metal with n impurities..1 Factrs cnsidered include cst, reducing strength needed, and by-prducts frmed.. a) In general, the mre active metal will be the ne with the lwer inizatin energy and/r the greater enthalpy f hydratin. In this way, net energy will be given ff when the mre active element lses electrns and the less active ne gains them, with the frmatin f ins f the mre active element and remval frm slutin f the ins f the less active element. An example is: Zn(s) + Cu + (aq) Cu(s) + Zn + (aq). b) This is similar t (a) except that hydratin enthalpies are nt invlved. An example is: Ca(l) + RbCl(l) Δ CaCl (l) + Rb(g).. In general, nnmetals are btained by xidatin and metals are btained by reductin..4 a) 6 b) 4 c) d) 1.5 a) Slag is a waste prduct f irn metallurgy frmed by the reactin: CaO(s) + SiO (s) CaSiO (l) In ther wrds, slag is a by-prduct f steel-making and cntains the impurity SiO. b) Pig irn, used t make cast irn prducts, is the impure prduct f irn metallurgy (cntaining 4% C) that is purified t steel. -

4 c) Steel refers t the prducts f irn metallurgy, specifically allys f irn cntaining small amunts f ther elements including 1 1.5% carbn. d) Basic-xygen prcess refers t the prcess used t purify pig irn t frm steel. The pig irn is melted and xygen gas under high pressure is passed thrugh the liquid metal. The xygen xidizes impurities t their xides, which then react with calcium xide t frm a liquid that is decanted. Mlten steel is left after the basic-xygen prcess..6 Pyrmetallurgy uses heat t btain the metal, electrmetallurgy uses electrical energy, and hydrmetallurgy uses reactins in aqueus slutin. a) Fe is prduced by a high-temperature prcess using C as the reductant. b) Na is prduced (with Cl and H ) electrchemically in a Dwns cell. c) Au is prduced in a hydrmetallurgical prcess using CN - as a cmplexing agent. d) Al is prduced electrchemically in the Hall-Herult prcess after a hydrmetallurgical step t disslve the re..7 Irn and nickel are mre easily xidized than cpper, s they are separated frm the cpper in the rasting step and cnversin t slag. In the electrrefining prcess, all three metals are xidized int slutin, but nly Cu + ins are reduced at the cathde t frm Cu(s)..8 Mlten crylite is a gd slvent fr Al O..9 a) Mlecules cntaining different istpes f a given atm react at different rates (heavier atms react mre slwly) if a bnd is brken t the atm in questin. b) H cmpunds exhibit a relatively large effect, since the mass rati f the istpes is larger fr H than fr any ther atm. c) It wuld be smaller fr C, since the mass rati f the istpes is smaller..0 Le Châtelier s principle says that the system shifts tward frmatin f K as the gaseus metal leaves the cell..1 a) Aqueus salt slutins are mixtures f ins and water. When tw half-reactins are pssible at an electrde, the ne with the mre psitive electrde ptential ccurs. In this case, the tw half-reactins are: M + + e M 0 E red =.05 V,.9 V, and.71 V fr Li +, K +, and Na +, respectively H O + e H + OH - E red = 0.4 V, with an vervltage f abut 1 V In all f these cases, it is energetically mre favrable t reduce H O t H than t reduce M + t M. b) The questin asks if Ca culd chemically reduce RbX, i.e., cnvert Rb + t Rb 0. In rder fr this t ccur, Ca 0 lses electrns (Ca 0 Ca + + e ) and each Rb + gains an electrn ( Rb + + e Rb 0 ). The reactin is written as fllws: RbX + Ca CaX + Rb where ΔH = IE 1 (Ca) + IE (Ca) IE 1 (Rb) = (40) = +99 kj/ml. Recall that Ca 0 acts as a reducing agent fr the Rb + in because it xidizes. The energy required t remve an electrn is the inizatin energy. It requires mre energy t inize calcium s electrns, s it seems unlikely that Ca 0 culd reduce Rb +. Based n values f IE and a psitive ΔH fr the frward reactin, it seems mre reasnable that Rb 0 wuld reduce Ca +. c) If the reactin is carried ut at a temperature greater than 688 C (the biling pint f rubidium), the prduct mixture will cntain gaseus Rb. This can be remved frm the reactin vessel, causing a shift in equilibrium t frm mre Rb prduct. If the reactin is carried ut between 688 C and 1484 C (b.p. fr Ca), then Ca remains in the mlten phase and remains separated frm gaseus Rb. d) The reactin f calcium with mlten CsX is written as fllws: CsX + Ca CaX + Cs where ΔH = IE 1 (Ca) + IE (Ca) IE 1 (Cs) = (76) = +98 kj/ml. This reactin is mre unfavrable than fr Rb, but Cs has a lwer biling pint f 671 C. If the reactin is carried ut between 671 C and 1484 C, then calcium can be used t separate gaseus Cs frm mlten CsX. -4

5 . a) Fr each mle f Na metal prduced, 0.5 mle f Cl is prduced. Calculate amunt f chlrine gas frm stichimetry, then use ideal gas law t find vlume f chlrine gas. NaCl(l) Na(l) + Cl (g) 10 g 1 ml Na 1mlCl Mles Cl = ( 1.0 kg Na) 1 kg = ml Cl (unrunded).99 g Na ml Na V = nrt / V = L atm ml Cl K ml K ( ) ( ) ( + ) ( 1.0 atm) = x 10 4 = 4.5 x 10 4 L b) Tw mles f electrns are passed thrugh the cell fr each mle f Cl prduced. - ml e C Culmbs = ( ml Cl ) - 1 ml Cl 1 ml e = x 10 8 = 1.0 x 10 8 Culmbs c) Current is charge per time with the amp unit equal t C / s. ( x 10 8 C 1 s ) 77 C = x 106 = 1.69 x 10 6 secnds. a) C (cke) b) Fuel (increasing temperature) c) CO d) Fe O (s) + CO(g) Fe O 4 (s) + CO (g) Fe O 4 (s) + CO(g) FeO(s) + CO (g) FeO(s) + CO(g) Fe(l) + CO (g).4 a) CaCO (s) Δ CaO(s) + CO (g) b) The lime (CaO) reacts with SiO present in the irn re t prduce CaSiO slag. The wrd flux cmes frm the Latin wrd meaning t flw. The liquid slag flws dwn t the bttm f the furnace and flats n the mre dense irn. c) CaO(s) + SiO (s) Δ CaSiO (l).5 a) Mg + is mre difficult t reduce than H O, s H (g) wuld be prduced instead f Mg metal. Cl (g) frms at the ande due t vervltage. b) The ΔH f fr MgCl (s) is kj/ml: Mg(s) + Cl (g) MgCl (s) + heat. High temperature favrs the reverse reactin (which is endthermic), s high temperatures favr the frmatin f magnesium metal and chlrine gas..6 a) Yes, it is near the bttm f the peridic table, thus it has a lw electrnegativity which allws it t adpt psitive xidatin states. b) IO (aq) + 5 HSO (aq) HSO 4 (aq) + SO 4 (aq) + H O(l) + I (s) Oxidizing agent: IO Reducing agent: HSO c) 0.78 ml% NaIO leaves ( ) ml% NaNO = 99. ml% NaNO 1kg 10 g 1 ml NaNO 0.78 ml% NaIO 1mlI 5.8gI ( 000. lb).05 lb 1 kg g NaNO ( 99.) ml% NaNO mlio 1 ml I = x 10 4 = 1.1 x 10 4 g I.7 a) Sulfur dixide is the reducing agent and is xidized t the +6 state (as sulfate in, S. O 4 ). b) The sulfate in frmed reacts as a base in the presence f acid t frm the hydrgen sulfate in. SO 4 (aq) + H + (aq) HSO 4 aq) -5

6 c) Skeletn redx equatin: SO (g) + H SeO (aq) Se(s) + HSO 4 (aq) Reductin half-reactin: H SeO (aq) Se(s) balance O and H H SeO (aq) + 4 H + (aq) Se(s) + H O(l) balance charge H SeO (aq) + 4 H + (aq) + 4 e Se(s) + H O(l) Oxidatin half-reactin: SO (g) HSO 4 aq) balance O and H SO (g) + H O(l) HSO 4 (aq) + H + (aq) balance charge SO (g) + H O(l) HSO 4 (aq) + H + (aq) + e Multiply xidatin half-reactin by and add the tw half-reactins. H SeO (aq) + SO (g) + H O(l) Se(s) + HSO 4 (aq) + H + (aq).8 The nly halgen capable f xidizing Cl is F, which is expensive and dangerus t handle. F als reacts with water, wasting the reagent..9 K SiF 6 (s) + 4 Al(s) 6 KF(s) + Si(s) + 4 AlF (s).40 mass % Fe = [(55.85 g Fe / ml) / ( g Fe O / ml)] x 100% = = 69.94% Fe mass % Fe = [(55.85 g Fe / ml) / (1.55 g Fe O 4 / ml)] x 100% = = 7.6% Fe mass % Fe = [1(55.85 g Fe / ml) / ( g FeS / ml)] x 100% = = 46.55% Fe.41 4 P(s) + 5 O (g) P 4 O 10 (s) P 4 O 10 (s) + 6 CaO(s) Ca (PO 4 ) (s).4 a) The xidatin state f cpper in Cu S is +1 (sulfur is ). The xidatin state f cpper in Cu O is +1 (xygen is ). The xidatin state f cpper in Cu is 0 (x state is always 0 in the elemental frm). b) The reducing agent is the species that is xidized. Sulfur changes xidatin state frm in Cu S t +4 in SO. Therefre, Cu S is the reducing agent, leaving Cu O as the xidizing agent..4 Al O is amphteric and will disslve in base: Al O (s) + NaOH(aq) + H O(l) Na[Al(OH) 4 ](aq) Fe O and TiO have n acidic prperties, s d nt disslve and can be remved..44 a) Use the surface area and thickness f the film t calculate its vlume. Multiply the vlume by the density t find the mass f the aluminum xide. Cnvert mass t mles f aluminum xide. The xidatin f aluminum invlves the lss f electrns fr each aluminum atm r 6 electrns fr each Al O cmpund frmed. Frm this, find the number f electrns prduced and multiply by Faraday s cnstant t find culmbs. 6 1cm.97 g AlO 1 ml AlO Mles Al O = (.5 m )(. x 10 m) 10 m cm g AlO =.8868 ml Al O 6 ml e C Culmbs = (.8868 ml AlO) 1 ml Al O 1 ml e = x 10 6 = 1. x 10 6 C b) Current in amps can be fund by dividing charge in culmbs by the time in secnds. [( x 10 6 C) / [15 min (60 s / min)] x (1 A / (C / s)) = 100 = 1. x 10 A.45 a) Wrk = ΔG = nfe Wrk = ( ml e ) (96485 C / ml e ) (1.4 V)[(J / C) / V] =.988 x 10 5 =.9 x 10 5 J (The in ml e is an exact number, and has n bearing n the significant figures.) b) Apparently the percent efficiency, nt the simple efficiency, is wanted. Efficiency = [(useful wrk dne) / (energy input)] x 100% Efficiency = [(.988 x 10 5 J) / (400 kj (10 J / 1 kj) )] x 100% = = 59.8% -6

7 c) Cst = ( 500. ml H ) 400 kj 1kW s 1 h $ ml H 1 kj 600 s kw h =. = $..46 a) H and CO b) The zelite acts as a mlecular filter t remve mlecules larger than H..47 Free energy, ΔG, is the measure f the ability f a reactin t prceed spntaneusly. The direct reductin f ZnS fllws the reactin: ZnS(s) + C(graphite) Zn(s) + CS (g) = [ ΔH f (Zn) + 1ΔH f (CS )] [ΔH f (ZnS) + 1ΔH f (C)] ΔG rxn ΔG rxn = [( ml) (0) + (1 ml) (66.9 kj/ml)] [( ml) ( 198 kj/ml) + (1 ml) (0)] = 46.9 = +46 kj Since ΔG is psitive, this reactin is nt spntaneus at standard-state cnditins. The stepwise reductin rxn invlves the cnversin f ZnS t ZnO, fllwed by the final reductin step: ZnO(s) + C(s) Zn(s) + CO (g) ΔG rxn = [ ΔH f (Zn) + 1ΔH f (CO )] [ΔH f (ZnO) + 1ΔH f (C)] = [( ml) (0) + (1 ml) ( 94.4 kj/ml)] [( ml) ( 18. kj/ml) + (1 ml) (0)] = +4.0 kj This reactin is als nt spntaneus, but the xide reactin is less unfavrable. ΔG rxn ΔG rxn ΔG rxn.48 The frmatin f sulfur trixide is very slw at rdinary temperatures. Increasing the temperature can speed up the reactin, but the reactin is exthermic, s increasing the temperature decreases the yield. Recall that yield, r the extent t which a reactin prceeds, is cntrlled by the thermdynamics f the reactin. Adding a catalyst increases the rate f the frmatin reactin but des nt impact the thermdynamics, s a lwer temperature can be used t enhance the yield. Catalysts are used in such a reactin t allw cntrl f bth rate and yield f the reactin..49 a) SO (g) + H SO 4 (l) H S O 7 (l) H S O 7 (l) + H O(l) H SO 4 (l) b) At high temperature, H O(g) catalyzes the plymerizatin f SO t (SO ) n, which frms a smke which makes pr cntact with the H O..50 All the steps in its prductin are exthermic, and the excess heat (in the frm f steam) can be sld prfitably..51 a) The chlr-alkali prcess yields Cl, H, and NaOH. b) The mercury-cell prcess yields higher purity NaOH, but prduces Hg-plluted waters that are discharged int the envirnment..5 a) The balanced equatin fr the xidatin f SO t SO : SO (g) + O (g) SO (g) ΔG = [( ml SO ) ( ΔG f f SO )] [( ml SO ) ( ΔG f f SO ) + (1 ml O ) ( ΔG f f O )] ΔG = [( ml SO ) ( 71 kj / ml SO )] [( ml SO ) ( 00. kj / ml SO ) + (1 ml O ) (0)] = = 14 kj The reactin is spntaneus at 5 C. b) The rate f the reactin is very slw at 5 C, s the reactin des nt prduce significant amunts f SO at rm temperature. c) Calculate standard state free energy at 500. C (77 K), using values fr ΔH and ΔS at 5 C with the assumptin that these values are cnstant with temperature. ΔH = [( ml SO ) ( ΔH f SO )] [( ml SO ) ( ΔH f SO ) + (1 ml O ) ( ΔH f O )] ΔH = [( ml SO ) ( 96 kj / ml SO )] [( ml SO ) ( 96.8 kj / ml SO ) + (1 ml O ) (0)] = = 198 kj -7

8 ΔS = [( ml SO ) (S f SO )] [( ml SO ) (S f SO ) + (1 ml O ) (S f O )] ΔS = [( ml SO ) (56.66 J/ml K)] [( ml SO ) (48.1 J/ml K ) + (1 ml O ) (05.0 J/ml K)] = = J/K ΔG 500 = ΔH TΔS = kj (( ) K) ( J/K) (1 kj / 10 J) = = 5 kj The reactin is spntaneus at 500 C since free energy is negative. d) The equilibrium cnstant at 500 C is smaller than the equilibrium cnstant at 5 C because the free energy at 500 C indicates the reactin des nt g as far t cmpletin as it des at 5 C. Equilibrium cnstants can be calculated frm ΔG = RT ln K. 10 J ( 14 kj) ΔG 1 kj At 5 C ln K = = = RT (8.14 J/mlK) ((7 + 5)K) K 5 = e = x 10 4 = 7.8 x J ( 5 kj) ΔG 1 kj At 500 C ln K = = = RT (8.14 J/mlK) (( )K) K 500 = e =.815 x 10 =.8 x 10 e) Temperature belw which the reactin is spntaneus at standard state can be calculated by setting ΔG equal t zer and using the enthalpy and entrpy values at 5 C t calculate the temperature. ΔH 198 kj T = = = x 10 = 1.05 x 10 K ΔS 1 kj J/K 10 J.5 This prblem deals with the stichimetry f electrlysis. The balanced xidatin half reactin fr the chlr-alkali prcess is given in the chapter: Cl (aq) Cl (g) + e Use the Faraday cnstant, F, (1 F = x 10 4 C / ml e ) and the fact that 1 ml f Cl prduces ml e r F, t cnvert culmbs t mles f Cl. C 4 Mass Cl = s 600 s 1 ml e 1 ml Cl g Cl 1 kg.05 lb ( x10 A) ( 8h) A 1 h C 1 ml Cl mle 10 g 1 kg = = 7 x 10 punds Cl.54 a) The prducts are kept separate because Cl reacts with NaOH(aq) t prduce NaOCl(aq) and NaCl(aq). b) The temperature at which the reactin is run determines the prduct frmed. c) Cl (g) + OH (aq) Cl (aq) + OCl (aq) + H O(l) Cl (g) + 6 OH (aq) 5 Cl (aq) + ClO (aq) + H O(l) Cl / OH is the mle rati in bth cases..55 SO (g) + O (g) SO (g) a) K p =.18 =.18 = ( PSO ) ( Ps ) ( P O 1 ( P ) O ) P SO = P SO because f the : rati f reactin cefficients. P O = = 0.14 atm O -8

9 b) K p =.18 = ( PSO ) ( Ps ) ( PO ) = ( 95) ( 5) ( PO ) P O = 11.5 = 1 x 10 atm O.56 Temperature belw which the reactin is spntaneus at standard state can be calculated by setting ΔG equal t zer and using the enthalpy and entrpy values at 5 C t calculate the temperature T = ΔH / ΔS. The enthalpy and entrpy changes must be determined first. It will simplify the prblem t cnvert the entrpy values t kj. a) MnO (s) Mn(s) + O (s) ΔH f = kj/ml S = J/ml K ΔH rxn = [1 ml Mn (0) + 1 ml O (0)] [1 ml MnO ( 50.9 kj/ml)] = 50.9 kj = [1 ml Mn (1.8 J/ml K) + 1 ml O (05.0 J/ml K)] [1 ml MnO (5.1 J/ml K)](1 kj / 10 J) = kj/k T = ΔH / ΔS = (50.9 kj) / (0.187 kj/k) = = 86 K b) MnO (s) + C(gr) Mn(s) + CO(g) ΔH f = kj/ml S = J/ml K ΔH rxn = [1 ml Mn (0) + ml CO ( kj/ml)] [1 ml MnO ( 50.9 kj/ml) + ml C (0)] = 99.9 kj = [1 ml Mn (1.8 J/ml K) + ml CO (197.5 J/ml K)] [1 ml MnO (5.1 J/ml K) + ml C (5.686 J/ml K)](1 kj / 10 J) = 0.6 kj/k T = ΔH / ΔS = (99.9 kj) / (0.6 kj/k) = = 87.8 K c) MnO (s) + C(gr) Mn(s) + CO (g) ΔH f = kj/ml S = J/ml K ΔH rxn = [1 ml Mn (0) + 1 ml CO (-9.5 kj/ml)] [1 ml MnO (-50.9 kj/ml) + 1 ml C (0)] = 17.4 kj = [1 ml Mn (1.8 J/ml K) + 1 ml CO (1.7 J/ml K)] [1 ml MnO (5.1 J/ml K) + 1 ml C (5.686 J/ml K)](1 kj / 10 J) = kj/k T = ΔH / ΔS = (17.4 kj) / ( kj/k) = = 68.4 K d) MnO (s) + Si(s) Mn(s) + SiO (s) ΔH f = kj/ml S = J/ml K ΔH rxn = [1 ml Mn (0) + 1 ml SiO ( kj/ml)] [1 ml MnO ( 50.9 kj/ml) + 1 ml Si (0)] = 90.0 kj = {[1 ml Mn (1.8 J/ml K) + 1 ml SiO (41.5 J/ml K)] [1 ml MnO (5.1 J/ml K) + 1 ml Si (18.0 J/ml K)]}(1 kj / 10 J) = 0.00 kj/k Since the enthalpy change is negative, and the entrpy change is psitive, this prcess is spntaneus at all temperatures..57 a) Because P 4 O 10 is a drying agent, water is incrprated in the frmatin f phsphric acid, H PO 4. P 4 O 10 (s) + 6 H O(l) 4 H PO 4 (l) b) First calculate the cncentratin (M) f the resulting phsphric acid slutin. 8.5 g P4O10 1 ml P4O10 4 ml HPO4 Mlarity = = M H PO 4 (unrunded) L 8.88 g P4O10 1 ml P4O10 Phsphric acid is a weak acid and nly partly dissciates t frm H O + in water, based n its K a. Since K a1 >> K a >> K a, assume that the H O + cntributed by the secnd and third dissciatin is negligible in cmparisn t the H O + cntributed by the first dissciatin. -9

10 H PO 4 (l) + H O(l) H O + (aq) + H PO 4 (aq) HO K a = 7. x 10 = HPO 4 [ x][ x] K a = 7. x 10 = x + HPO 4 ( ) The prblem will need t be slved as a quadratic. x = K a ( x) = (7. x 10 ) ( x) = x x 10 x x + 7. x 10 x x 10 = 0 a = 1 b = 7. x 10 c = x 10 ( ) ()( ) 7. x 10 ± 7. x x 10 x = 1 () x = x 10 M H O + (unrunded) ph = lg [H + ] = lg ( x 10 ) = = a) D + (aq), OD (aq) and traces f CH O (aq) and H + (aq). Eventually, HDO, H O, and CH OD wuld frm. b) D O(l) D + (aq) + OD (aq) CH OH(aq) + OD (aq) CH O (aq) + HDO(l) CH O (aq) + D O(l) CH OD(l) + OD (aq) HDO(l) D + (aq) + OH (aq) HDO(l) + OH (aq) H O(l) + OD (aq).59 The reactin taking place in the blast furnace t prduce irn is: Fe O (s) + CO(g) Fe(s) + CO (g) (t = tn = 000 lbs, and nt T = metric tn = 1000 kg) a) Three mles f carbn dixide are prduced fr every mles f Fe. 000 lb 1 kg 10 g 1 ml Fe ml CO g CO Mass = ( 8400.t Fe) 1 t.05 lb 1 kg g Fe ml Fe 1 ml CO = x 10 9 = x 10 9 g CO b) Cmbustin reactin fr ctane: C 8 H 18 (l) + 5 O 16 CO (g) + 18 H O(l) gal 4 qt 1 L 1 ml 0.74 g 1mlC8H18 16 ml CO g CO ( 1.0 x 10 auts) 1 aut 1 gal qt 10 L ml 114. g C8H18 ml C8H18 1 ml CO = x = 4. x g CO The CO prductin by autmbiles is much greater than that frm steel prductin..60 a) Step 1 Cl (g) + CH =CH (g) CH ClCH Cl(g) Step CH ClCH Cl(g) CH =CHCl(g) + HCl(g) b) Cl (g) + CH =CH (g) CH =CHCl(g) + HCl(g) c) Additin reactin d) Eliminatin reactin 1 ml CH = CHCl 6.49 g CH = CHCl e) ( 0.45 ml Cl ) = = 8 g CH =CHCl 1 ml Cl 1 ml CH = CHCl -10

11 .61 a) Step f Islatin and Uses f Magnesium describes this reactin. Sufficient OH must be added t precipitate Mg(OH). The necessary amunt f OH can be determined frm the K sp f Mg(OH). Mg(OH) (s) Mg + (aq) + OH (aq) K sp = 6. x = [Mg + ][OH ] [OH ] = [(6. x ) / (0.05)] 1/ = x 10 4 = 1.1 x 10 4 M Thus, if [OH ] > 1.1 x 10 4 M (i.e., if ph > 10.04), Mg(OH) will precipitate. b) The amunt f OH that a saturated slutin f Ca(OH) prvides is dependent n its K sp. Ca(OH) (s) Ca + (aq) + OH (aq) K sp = 6.5 x 10 6 = [Ca + ][OH ] = (x) (x) = 4x x = ; [OH ] = x = M (unrunded) Substitute this amunt int the K sp expressin fr Mg(OH) t determine hw much Mg + reacts with this amunt f OH. K sp (Mg(OH) ) = 6. x = [Mg + ][OH ] [Mg + ] = (6. x ) / ( ) = x 10 6 M (unrunded) This cncentratin is the amunt f the riginal 0.05 M Mg + that was nt precipitated. The percent precipitated is the difference between the remaining [Mg + ] and the initial [Mg + ] divided by the initial cncentratin. Fractin Mg + remaining = ( x 10 6 M) / (0.05 M) =.1919 x 10 5 (unrunded) Mg + precipitated = x 10 5 = = 1 (t the limit f the significant figures, all the magnesium has precipitated.).6 a) Reactin () is a disprprtinatin reactin in which the N changes frm a +4 xidatin state in NO t +5 in HNO and + in NO. b) 4 NH (g) + 5 O (g) 4 NO(g) + 6 H O(g) 6 NO(g) + O (g) 6 NO (g) 6 NO (g) + H O(g) 4 HNO (aq) + NO(g) Overall: 4 NH (g) + 8 O (g) 4 HNO (aq) + 4 H O(g) r NH (g) + O (g) HNO (aq) + H O(g) c ) ΔH rxn = [1 ml HNO ( kj/ml) + 1 ml H O ( kj/ml)] [1 ml NH (-45.9 kj/ml) + ml O (0 kj/ml)] = = 40.5 kj.6 It will be necessary t calculate K p frm ΔG. The equatin ΔG = ΔH TΔS will be used t calculate ΔG at the tw temperatures. Then use ΔG = RT ln K t determine K. Subscripts indicate the temperature, and (1) r () indicate the reactin. Fr the first reactin (1): = [4 ml NO (90.9 kj/ml) + 6 ml H O ( kj/ml)] [4 ml NH ( 45.9 kj/ml) ΔH rxn + 5 ml O (0 kj/ml)] = kj (unrunded) = [4 ml NO (10.65 J/K ml) + 6 ml H O (188.7 J/K ml)] [4 ml NH (19 J/K ml) + 5 ml O (05.0 J/K ml)](1 kj / 10 J) = kj/k (unrunded) ΔG = ΔH TΔS ΔG 5 = kj [(7 + 5)K]( kj/k) = kj (unrunded) ΔG 900 = kj [( )K]( kj/k) = kj (unrunded) Fr the secnd reactin (): = [ ml N (0 kj/ml) + 6 ml H O ( kj/ml)] ΔH rxn [4 ml NH ( 45.9 kj/ml) + ml O (0 kj/ml)] = kj (unrunded) = [ ml N (191.5 J/K ml) + 6 ml H O (188.7 J/K ml)] [4 ml NH (19 J/K ml) + ml O (05.0 J/K ml)]}(1 kj / 10 J) = 0.18 J/K (unrunded) ΔG = ΔH TΔS ΔG 5 = kj [(7 + 5)K](0.18 kj/k) = kj (unrunded) ΔG 900 = kj [( )K](0.18 kj/k) = kj (unrunded) -11

12 ( ) ( ) ( ) ΔG kj/ml 10 J a) ln [K 5 (1)] = = RT 8.14 J/ml K ( 7 5 K) = (unrunded) + 1kJ K 5 (1) = 1.8 x = 1 x ΔG ( kj/ml) 10 J ln [K 5 ()] = = RT ( 8.14 J/ml K) (( 7 5) K) = (unrunded) + 1kJ K 5 () = x 10 8 = 7 x 10 8 ΔG ( kj/ml) 10 J b) ln [K 900 (1)] = = RT ( 8.14 J/ ml K) (( 7 900) K) = (unrunded) + 1kJ K 900 (1) = x = 4.6 x ( ) ( ) ( ) ΔG kj/ml 10 J ln [K 900 ()] = = RT 8.14 J/ ml K ( K) = (unrunded) + 1kJ K 900 () = 1.84 x 10 6 = 1.4 x c) Cst ($) Pt = ( 1.01 x 10 t HNO ) 175 mg Pt 10 g 1 kg.15 Try Oz $160 1 t HNO 1 mg 10 g 1 kg 1 Try Oz = x 10 7 = $ 7. x d) ( 1.01 x 10 t HNO ) 0.97 g P.64 a) mass % P = 1.06 g (NH ) HPO 1 t HNO 100% 1 mg 1 kg 1 try z = $ x 10 7 = $5.16 x mg Pt 7% 10 g 1 kg.15 try z $ g 4 4 x 100% = =.45% P in (NH 4 ) HPO g P mass % P = x100% = = 6.9% P in NH 4 (H PO 4 ) g NH4HPO4 NH 4 (H PO 4 ) has the higher % P. b) It might be less expensive, yu might prefer ne with a higher % N, r yu might prefer a less acidic cmpund..65 Balanced reactins: FeTiO (s) + 7 Cl (g) + 6 C(s) TiCl 4 (g) + FeCl (l) + 6 CO(g) TiCl 4 (g) + Mg(l) Ti(s) + MgCl (l) 10 kg 10 g 1mlFeTiO 1 ml Ti g Ti 84% 9% Mass Ti = ( 1.5 t FeTiO ) 1 t 1 kg g FeTiO 1 ml FeTiO 1 ml Ti 100% 100% = x 10 6 = 5. x 10 6 g Ti.66 a) Balance the tw steps and then add them. (1) 16 H S(g) + 16 O (g) S 8 (g) + 8 SO (g) + 16 H O(g) () 16 H S(g) + 8 SO (g) S 8 (g) + 16 H O(g) H S(g) + 16 O (g) 4 S 8 (g) + H O(g) The cefficients in the verall equatin can be reduced since all are divisible by 4. 8 H S(g) + 4 O (g) S 8 (g) + 8 H O(g) b) Replace xygen with chlrine: 8 H S(g) + 8 Cl (g) S 8 (g) + 16 HCl(g) Calculate ΔG t determine if this reactin is thermdynamically pssible. ΔG = [(1 ml S 8 ) (49.1 kj/ml) + (16 ml HCl) ( 95.0 kj/ml)] [(8 ml H S) ( kj/ml) + (8 ml Cl ) (0)] = = 11 kj The reactin is spntaneus. -1

13 c) Oxygen is readily available frm the air, s chlrine is a mre expensive reactant. In additin, water is less crrsive than HCl as a prduct..67 (1) H O(l) + FeS (s) + 7 O (g) Fe + (aq) + 4 SO 4 (aq) + 4 H + (aq) Increase acidity. () 4 H + (aq) + 4 Fe + (aq) + O (g) 4 Fe + (aq) + H O(l) () Fe + (aq) + H O(l) Fe(OH) (s) + H + (aq) Increase acidity. (4) 8 H O(l) + FeS (s) + 14 Fe + (aq) 15 Fe + (aq) + SO 4 (aq) + 16 H + (aq) Increase acidity. price 8 $.91 x 10 Br 0.6 mi 1 km 10 m 1 cm 1 ml 1 L g Br.68 a) = gram Br mi 1 km 10 m 1 cm 1 ml 10 L g Br g Br = x 10 4 = $ 7. x 10 4 / g Br g Br 1 ml Br 90.0% 1 ml AgNO g AgNO b) Mass f AgNO = ( 1.00 L) 1 L g 100% 1mlBr 1 ml AgNO = = 0.1 g AgNO c) Mlarity f Br remaining after 90.0% precipitated = 10.0% g Br 1 ml Br 100% 1 L = 8.15 x g Br 5 M (unrunded) K sp (AgBr) = 5.0 x 10 1 = [Ag + ][Br ] = [Ag + ][8.15 x 10 5 M] [Ag + ] = x 10 9 M (unrunded) K sp (AgCl) = 1.8 x = [Ag + ][Cl ] = [ x 10 9 M][Cl ] [Cl ] =.986 x 10 M (unrunded) The chlride cncentratin must be less than.9 x 10 M t ensure that n AgCl precipitates. d) If the [Cl ] = 0. M then it is nt practical t try t precipitate AgBr, because the Ag + ins will first precipitate ut with the Cl ins..69 The carbn fits int interstitial psitins. The cell size will increase slightly t accmmdate the carbns atms. The increase in size is assumed t be negligible. The mass f the carbn added depends upn the carbn in the unit cell. Ferrite: 7.86 g 7.86 g 0.018% + cm cm 100% = = 7.86 g/cm Austenite: 7.40 g 7.40 g.08% + cm cm 100% = = 7.55 g/cm.70 (1) N (g) + O (g) NO (g) ΔG =[ ( ml) (51 kj/ml)] [(1 ml) (0 kj/ml) + ( ml) (0 kj/ml)] = 10 kj (unfavrable) () NO (g) + H O(l) HNO (aq) + NO(g) ΔG = [( ml) ( kj/ml) + (1 ml) (86.60 kj/ml)] [( ml) (51 kj/ml) + (1 ml) ( 7.19 kj/ml)] = = 50. kj (favrable) () NO(g) + O (g) NO (g) ΔG = [( ml) (51 kj/ml)] [( ml) (86.60 kj/ml) + (1 ml) (0 kj/ml)] = 71. = 71 kj (favrable) (verall) N (g) + 6 O (g) + H O(l) 4 HNO (aq) + NO(g) ΔG = [4 ml ( kj/ml) + ( ml) (86.60 kj/ml]) [( ml) (0 kj/ml) + (6 ml) (0 kj/ml) + ( ml) ( 7.19 kj/ml)] = = 05.6 kj Overall, the reactin is thermdynamically unfavrable, due t the unfavrability f the first step. -1

14 .71 a) Decide what is reduced and what is xidized f the tw ins present in mlten NaOH. Then, write halfreactins and balance them. At the cathde, sdium ins are reduced: Na + (l) + e Na(l) At the ande, hydrxide ins are xidized: 4 OH (l) O (g) + H O(g) + 4 e b) The verall cell reactin is 4 Na + (l) + 4 OH (l) 4 Na(l) + O (g) + H O(g) and the reactin between sdium metal and water is Na(l) + H O(g) NaOH(l) + H (g). Fr each mle f water prduced, mles f sdium are prduced, but fr each mle f water that reacts nly 1 mle f sdium reacts. S, the maximum amunt f sdium that reacts with water is half f the sdium prduced. In the balanced cell reactin, 4 mles f electrns are transferred fr 4 mles f sdium. Since 1/ f the Na reacts with H O, the maximum efficiency is 1/ ml Na/ml e, r 50%..7 a) 4 Au(s) + O (g) + H O(l) 4 Au + (aq) + 4 OH (aq) ph = 1.55, poh = ph = = 0.45 [OH ] = = M (unrunded) E = E O E Au = 0.40V 1.68V = 1.8V E = E n lg Q = E V n + Au OH lg Pxygen [ 0.50 ] [ ] lg 4 4 E = 1.8 V V 4 4 = = 1.5 V This reactin is nt spntaneus (the vltage is negative). b) Qualitatively, the additin f CN will lwer the cncentratin f free Au + by frming Au(CN). This will make [Au + ] much lwer than 1 M, making the secnd term in the Nernst equatin larger in magnitude. If [Au + ] is lwered sufficiently, the secnd term ffsets the first and E cell becmes favrable..7 a) Nitric xide destrys zne by frming nitrgen dixide and xygen. NO(g) + O (g) NO (g) + O (g) NO (g) + O (g) NO(g) + O (g) (reverse) b) The rder translates t the expnent in the rate law, and the rate is dependent n the cncentratin f the reactants. Rate f = k f [NO][O ] Rate r = k r [NO ][O ] c) Use ΔG = ΔH TΔS because the reactin des nt take place at 98 K. ΔH rxn = [(1 ml NO ) (. kj/ml) + (1 ml O ) (0)] [(1 ml NO) (90.9 kj/ml) + (1 ml O ) (14 kj/ml)] ΔH rxn = kj (unrunded) = [(1 ml NO ) (9.9 J/ml K) + (1 ml O ) (05.0 J/ml K)] [(1 ml NO) (10.65 J/ml K) + (1 ml O ) (8.8 J/ml K)] = 4.57 J/K (unrunded) ΔG 80 = ΔH T = ( kj) (80. K) ( 4.57 J/K) (kj / 10 rxn J) = = 199 kj. d) Assume that the reactin reaches equilibrium at 80. K. At equilibrium, the frward rate equals the reverse rate and their rati equals the equilibrium cnstant: k f / k r = K eq. Bth ΔG and K eq express the extent t which a reactin prceeds and are related by the equatin ΔG = RT ln K eq kj/ ml 10 J ln K = ΔG / RT = ( 8.14 J/ml K)( 80.K) = (unrunded) 1 kj K = x 10 7 = 1. x 10 7 Therefre, the rati f rate cnstants, k f / k r, is 1. x 10 7 and the frward rate is much faster than the reverse rate. -14

15 .74 a) The reactin fr the fixatin f carbn dixide includes CO and H O as reactants and (CH O) n and O as prducts. The balanced equatin is: n CO (g) + n H O(l) (CH O) n (s) + n O (g) b) The mles f carbn dixide fixed equal the mles f O prduced. 48 g CO 1 ml CO 1 ml O Mle O / day = = ml O / day day g CO 1 ml CO T find the vlume f xygen, use the ideal gas equatin. Temperature must be cnverted t K. 5 T = T ( F ) = 5 T ( 78. ) = K V = nrt / P = ( ml O )( L atm/ ml K)( K) ( 1.0 atm) c) The mles f air cntaining ml f CO is: 48 g CO 1 ml CO 100 ml % air Mle CO / day = day g CO 0.05 ml % CO = = 7 L = x 10 ml air / day ( ) ( )( V = nrt / P = x 10 ml air L atm/ ml K K ( 1.0 atm) ) = x 10 4 = 7.6 x 10 4 L.75 This prblem requires a series f cnversin steps: Cncentratin = 10. kg(nh 4) SO 4 10 g 1 ml (NH 4) SO4 ml NO 7% 6.01 g NO 1 mg 10 m 1 kg 1.15 g (NH m 4) SO 4 1 ml (NH 4) SO 4 100% 1 ml NO 1 L 10 g = = 7 mg/l This assumes the plants in the field absrb nne f the fertilizer..76 a) Se (aq) Se(s) + e b) H O(l) + SeO (aq) SeO 4 (aq) + H O + (aq) + e c) 6 e + 8 H O + (aq) + SeO 4 (aq) Se(s) + 1 H O(l).77 The balanced chemical equatin fr this reactin is: 6 HF(g) + Al(OH) (s) + NaOH(aq) Na AlF 6 (aq) + 6 H O(l) Determine which ne f the three reactants is the limiting reactant. 1) The amunt f Al(OH) requires cnversin f kg t g, and then determinatin f mles f crylite frmed. 10 g 1mlAl(OH) 1mlNaAlF6 ml Na AlF 6 = ( 65 kg Al(OH) ) 1 kg g Al(OH) 1 ml Al(OH) = x 10 ml Na AlF 6 (unrunded) ) The amunt f NaOH requires cnversin frm amunt f slutin t g, then determinatin f mles f crylite frmed. 1 L 1 ml 1.5 g 50.0% NaOH 1 ml NaOH 1mlNaAlF6 ml Na AlF 6 = ( 1.0 m ) 10 m 10 L ml 100% g NaOH ml NaOH = x 10 ml Na AlF 6 (unrunded) ) The amunt f HF requires cnversin frm vlume t mles using the ideal gas law, and then determinatin f mles f crylite frmed. n = PV/RT -15

16 ( )( ) 05kPa 65m ml Na AlF 6 = 10 Pa 1 atm 1 L 1mlNaAlF6 5 ( L/ ml K) (( ) K) 1kPa x 10 Pa 10 m 6mlHF = x 10 ml Na AlF 6 (unrunded) Because the 65 m f HF prduces the smallest amunt f Na AlF 6, it is the limiting reactant. Cnvert mles Na AlF 6 t grams Na AlF 6, using the mlar mass, cnvert t kg, and multiply by 95.6% t determine the final yield g NaAlF6 1 kg 95.6% mass crylite = ( x 10 ml NaAlF6) 1mlNa AlF 6 10 g 100% = = 891 kg Na AlF 6.78 a) The rate f effusin is inversely prprtinal t the square rt f the mlar mass f the mlecule. rateh MD = rated M = 4.00 g/ml = (unrunded) H.016 g/ml The time fr D t effuse is 1.41 times greater than that fr H. Time D = ( ) (16.5 min) =.417 =. min b) Set x equal t the number f effusin steps. The rati f ml H t ml D is 99:1. Set up the equatin t slve fr x: 99 / 1 = ( ) x. When slving fr an expnent, take the lg f bth sides. lg(99) = lg( ) x Remember that lg(a b ) = b lg(a) lg(99) = x lg( ) x = (unrunded) T separate H and D t 99% purity requires 14 effusin steps..79 a) The equilibrium cnstant is large, the rate cnstant is small, and E a is large. b) CO(g) C(graphite) + CO (g) c) ΔG = [(1 ml) (0 kj/ml) + (1 ml) ( 94.4 kj/ml)] [ ( ml) ( 17. kj/ml]) = 10.0 kj ΔG = RT ln K p ( 10.0 kj) 10 J ln K p = 8.14 K/ml K 98 K = (unrunded) 1 kj ( )( ) K p = x 10 1 = 1.1 x 10 1 d) K p = K c (RT) Δngas K c =K p / (RT) Δngas = 1 ( x 10 ) ( )( ) = x 10 =.6 x a) Use the stichimetric relatinships fund in the balanced chemical equatin t find mass f Al O. Assume that 1 metric tn Al is an exact number. Al O (in Na AlF 6 ) + C(gr) 4 Al(l) + CO (g) 10 kg 10 g 1 ml Al ml AlO g AlO 1 kg 1 t mass Al O = ( 1tAl) 1 t 1 kg 6.98 g Al 4 ml Al 1 ml AlO 10 g 10 kg = = metric tns Al O Therefre, tns f Al O are cnsumed in the prductin f 1 tn f pure Al. b) Use a rati f ml C: 4 ml Al t find mass f graphite cnsumed. 10 kg 10 g 1 ml Al ml C 1.01 g C 1 kg 1 t mass C = ( 1tAl) 1 t 1 kg 6.98 g Al 4 ml Al 1 ml C 10 g 10 kg = = 0.9 tns C Therefre, 0.9 tns f C are cnsumed in the prductin f 1 tn f pure Al, assuming 100% efficiency. -16

17 c) The percent yield with respect t Al O is 100% because the actual plant requirement f 1.89 tns Al O equals the theretical amunt calculated in part (a). d) The amunt f graphite used in reality t prduce 1 tn f Al is greater than the amunt calculated in (b). In ther wrds, a 100% efficient reactin takes nly 0.9 tns f graphite t prduce a tn f Al, whereas real prductin requires mre graphite and is less than 100% efficient. Calculate the efficiency using a simple rati: (0.45 t) (x) = ( t) (100%) x = = 74% e) Fr every 4 mles f Al prduced, mles f CO are prduced. 10 kg 10 g 1 ml Al mlco mles C = ( 1tAl) 1 t 1 kg =.7798 x 10 4 ml CO (unrunded) 6.98 g Al 4 ml Al The prblem states that 1 atm is exact. Use the ideal gas law t calculate vlume, given mles, temperature and pressure. 4 (.7798 x 10 ml CO )( )(( ) ) L atm/ml K K 10 m V = nrt / P = 1atm 1L = x 10 =.81 x 10 m.81 a) The reactins are: 4 FeCr O 4 (s) + 8 Na CO (aq) + 7 O (g) 8 Na CrO 4 (aq) + Fe O (s) + 8 CO (g) Na CrO 4 (aq) + H O + (aq) Na Cr O 7 (s) + H O(l) + Na + (aq) Na Cr O 7 (s) + C(s) Cr O (s) + Na CO (s) + CO(g) Cr O (s) + Al(s) Cr(s) + Al O (s) b) mass Cr = 7 10 kg 10 g 1mlFeCrO 4 ml Cr 5.00 g Cr 1 kg ( 1.5 x 10 t FeCrO 4) 1 t 1 kg.85 g FeCrO 4 1 ml FeCrO 4 1 ml Cr 10 g = x 10 9 = 7.0 x 10 9 kg Cr.8 a) HDO + H HD + H O b) At equilibrium, there are mre reactants than prducts. This is a reactant-favred reactin s the equilibrium cnstant K must be less than 1. [ H ][ ] O HD [ 0.0][ 0.0] c) K = = HDO H = 0.44 [ ][ ] [ ][ ].8 K sp (NiS) = 1.1 x = [Ni + ][HS ][OH ] = (0.10)[HS ][OH ], s if [HS ][OH ] < 1.1 x 10 17, K sp will nt be exceeded and NiS will nt precipitate. K sp (CuS) = 8 x 10 4 = [Cu + ][HS ][OH ] = (0.10)[HS ][OH ], s if [HS ][OH ] > 8 x 10, K sp will be exceeded and CuS will precipitate. HS + H K a1 = 9 x 10 8 O HS + H = O = HS 0.10 [ ] [HS ][H O + ] = 9 x 10 9, r [HS ] = 9 x 10 9 M / [H O + ] K w = [H O + ][OH ] r [OH ] = K w / [H O + ] s [HS ][OH ] = (9 x 10 9 ) (1.0 x ) / [H O + ] = (9 x 10 ) / [H O + ] [HS ][OH ] < 1.1 x [H O + ] > x 10 M and ph <.5 [HS ][OH ] > 8 x 10 [H O + ] < 1 x 10 5 M and ph > 5.0 S, maintaining the ph between 5.0 and.5 wuld allw CuS t precipitate but prevent precipitatin f NiS. -17

18 .84 a) mass re = (.0 x 10 lb Cu) 5 1 kg 100%.05 lb 0.5% =.681 x 107 =.6 x 10 7 kg re 0.5% Cu g FeCuS b) mass% chalcpyrite = ( 100% 100% ) = 0.70 = 0.7% chalcpyrite 6.55 g Cu.85 Acid rain increases the leaching f PO 4 int the grundwater, due t the prtnatin f PO 4 t frm HPO 4 and H PO 4. As shwn in calculatins a) and b), slubility increases frm 6.4 x 10 7 M (in pure water) t 1.1 x 10 M (in acidic rainwater). a) Slubility f a salt can be calculated frm its K sp. K sp fr Ca (PO 4 ) is 1. x Ca (PO 4 ) (s) Ca + (aq) + PO 4 (aq) Initial 0 0 Change x +x +x Equilibrium x x K sp = [Ca + ] [PO 4 ] = ( x) ( x) = 108 x 5 = 1. x x = x 10 7 = 6.4 x 10 7 M Ca (PO 4 ) b) Phsphate is derived frm a weak acid, s the ph f the slutin impacts exactly where the acid-base equilibrium lies. Phsphate can gain ne H + t frm HPO - 4. Gaining anther H + gives H PO 4 and a last H + added gives H PO 4. The K a values fr phsphric acid are K a1 = 7. x 10, K a = 6. x 10 8, K a = 4. x T find the ratis f the varius frms f phsphate, use the equilibrium expressins and the cncentratin f H + : [H O + ] = =.16 x 10 5 M (unrunded). H PO 4 (aq) + H O(l) H PO 4 (aq) + H O + (aq) + HPO 4 HO K a1 = HPO 4 HPO 4 Ka1 7. x 10 = = = 7.70 (unrunded) HPO 4 HO x 10 H PO 4 (aq) + H O(l) HPO 4 (aq) + H O + (aq) + HPO 4 HO K a = HPO 4 HPO 8 4 Ka 6. x 10 = = = x 10 (unrunded) HPO 4 HO x 10 HPO 4 (aq) + H O(l) PO 4 (aq) + H O + (aq) + PO 4 HO K a = HPO 4 PO 1 4 Ka 4. x 10 = = = 1.87 x 10 8 (unrunded) HPO 4 HO x 10-18

19 Frm the ratis, the cncentratin f H PO 4 is at least 100 times mre than the cncentratin f any ther species, s assume that dihydrgen phsphate is the dminant species and find the value fr [PO 4 ]. PO 4 HPO = 1.87 x 10 8 and 4 = x 10 HPO 4 HPO 4 [PO 4 ] = (1.87 x 10 8 )[HPO 4 ] and [HPO 4 ] = (1.994 x 10 )[H PO 4 ] [PO 4 ] = (1.87 x 10 8 )[HPO 4 ] = (1.87 x 10 8 ) (1.994 x 10 )[H PO 4 ]. [PO 4 ] = (.6464 x 10 11)( )[H PO 4 ]) Substituting this int the K sp expressin gives K sp = [Ca + ] [PO 4 ] = K sp = [Ca + ] {(.6464 x )[H PO 4 ]} Rearranging gives K sp / (.6464 x ) = [Ca + ] [H PO 4 ] The cncentratin f calcium ins is still represented as x and the cncentratin f dihydrgen phsphate in as x, since each H PO 4 cmes frm ne PO 4. K sp / (.6464 x ) = ( x) ( x) (1. x 10 9 ) / (.6464 x ) = 108 x 5 x = x 10 = 1.1 x 10 M.86 a) The mle % f xygen missing may be estimated frm the discrepancy between the actual and ideal frmula. ( ) ml O ml % O missing = x 100% = 1.00%.000 ml O b) The mlar mass is determined like a nrmal mlar mass except that the quantity f xygen is nt an integer value. mlar mass = 1 ml Pb (07. g/ml) ml O (16.00 g/ml) = 8.88 = 8.9 g/ml.87 Using ΔG values: H O(l) H (g) + 1/ O (g) ΔG = kj/ml H S(g) H (g) + 1/8 S 8 (s) ΔG = + kj/ml Bth these prcesses require energy t prceed, but the secnd requires less energy and wuld be mre favrable (r less unfavrable) based n this criterin..88 Density Silver = 4 Ag atms 1 ml Ag g Ag unit cell 6.0 x 10 Ag atms 1 ml Ag = = g/cm ( ) pm 10 m 1 cm unit cell 1 pm 10 m The calculatin will nw be repeated replacing the atmic mass f silver with the atmic mass f sterling silver. The atmic mass f sterling silver is simply a weighted average f the masses f the atms present (Ag and Cu): ( ) g % 6.55 g 7.5% Atmic mass f sterling silver = ml + 100% ml 100% = g/ml (unrunded) Density Sterling Silver = 4 Ag atms 1 ml Ag g"ag" unit cell 6.0 x 10 Ag atms 1 ml Ag = = 10. g/cm ( ) pm 10 m 1 cm unit cell 1 pm 10 m -19

20 .89 a) Mles Ti = ( 5.98 x 10 kg) b) Mass ilmenite = ( 5.98 x 10 kg) 4 10 g 0.05% 1 ml Ti 1 kg 100% = 6.4 x 10 = 6 x 10 ml Ti g Ti = x 10 4 = 5 x 10 4 g FeTiO c) Years = ( 5.98 x10 4 kg).90 a) NaCl(s) + H SO 4 (aq) Na a SO 4 (aq) + HCl(g) Na SO 4 (s) + C(s) Na S(s) + CO (g) Na S(s) + CaCO (s) Na CO (s) + CaS(s) Ttal: 1 kg 100% g Ti 100% 1 ml Ti 1 ml FeTiO 4 10 g 0.05% 1 ml Ti 50% 1 ml FeTiO g FeTiO 0.05%.05 lbs 1 tn year 5 100% 1 kg 000 lb 1.00 x 10 tns =.96 x 101 = x 10 1 years CaCO (s) + Na SO 4 (s) + C(s) + NaCl(s) + H SO 4 (aq) Na a SO 4 (aq) + HCl(g) + CO (g) + Na CO (s) + CaS(s) b) Ntice that Na SO 4 (aq) and Na SO 4 (s) are different substances, and, as such, will nt cancel in this Hess s Law calculatin. The values fr Na SO 4 (aq) and Na SO 4 (s) are nt in the Appendix s they must be fund frm anther surce. The ther enthalpies f frmatin are either given in the prblem r are given in the Appendix. ΔH rxn = ΣmΔH f (prducts) - ΣnΔH f (reactants) Δ H rxn = [(1 ml) ΔH f (Na a SO 4 (aq)) + ( ml) ΔH f (HCl(g)) + ( ml) ΔH f (CO (g)) + (1 ml) ΔH f (Na CO (s)) + (1 ml) ΔH (CaS(s))] [(1 ml) ΔH f (CaCO (s)) + (1 ml) ΔH f (Na SO 4 (s)) + ( ml) ΔH (C(s)) + f f ( ml) ΔH f (NaCl(s)) + (1 ml) ΔH f (H SO 4 (aq))] 51.8 kj/ml = [1 ml( kj/ml) + ml( 9.1 kj/ml) + ml( 9.5 kj/ml) + 1 ml( kj/ml) + 1 ml( ΔH f (CaS(s))] [1 ml( kj/ml) + 1 ml( kj/ml) + ml(0 kj/ml) + ml( kj/ml) + 1 ml( kj/ml)] ΔH f CaS(s) = = kj c) The reactin is very endthermic, therefre it is prbably nt spntaneus even thugh the entrpy change may be expected t be psitive. It will take a ΔG determinatin t knw fr sure. 1mlNaCl 1 ml NaCO g NaCO 7% d) ( 50. g NaCl) g NaCl ml NaCl 1 ml NaCO 100% = = 165 g Na CO.91 a) Fr CaCO : CaCO (s) CaO(s) + CO (g) ΔH rxn = Σm ΔH f (prducts) Σn ΔH f (reactants) ΔH rxn = [(1 ml) ΔH f (CaO(s)) + (1 ml) ΔH f (CO (g))] [(1 ml) ΔH f (CaCO (s))] ΔH rxn = [1 ml ( 65.1 kj/ml) + 1 ml( 9.5 kj/ml)] [1 ml ( kj/ml)] ΔH rxn = 178. kj = ΣmS (prducts) - ΣnS (reactants) = [(1 ml)s (CaO(s)) + (1 ml)s (CO (g))] [( 1 ml)s (CaCO (s))] = [1 ml (8. J/ml K) + 1 ml (1.7 J/ml K)] [1 ml (9.9 J/ml K)] = J/K Set: ΔG = ΔH TΔS = 0 ΔH = TΔS 178. kj 10 J T = ΔH / ΔS = J/K 1 kj = = 111 K fr CaCO -0

21 Fr MgCO : MgCO (s) MgO(s) + CO (g) ΔH rxn = Σm ΔH f (prducts) Σn ΔH f (reactants) ΔH rxn ΔH rxn ΔH rxn = [1 ml) ΔH f (MgO(s)) + (1 ml) ΔH f (CO (g))] [(1 ml) ΔH f (MgCO (s))] = [1 ml ( 601. kj/ml) + 1 ml ( 9.5 kj/ml)] [1 ml ( 111 kj/ml)] = 117. kj (unrunded) = ΣmS (prducts) ΣnS (reactants) = [(1 ml) S (MgO(s)) + (1 ml) S (CO (g))] [(1 ml)s (MgCO (s))] = [1 ml (6.9 J/ml K) + 1 ml (1.7 J/ml K)] [1 ml (65.86 J/ml K)] = J/K (unrunded) Set ΔG = ΔH TΔS = 0 ΔH = TΔS 117. kj 10 J T = ΔH / ΔS = J/K 1 kj = = 671 K fr CaCO b) CaO(s) + SiO (s) CaSiO (s) 6 50 kg slag 10 g 1 ml slag 1 ml CaCO g CaCO 1 kg 84% Mass = (.7 x 10 tns) tn 1 kg g slag 1 ml slag 1 ml CaCO 10 g 100% = x 10 7 = 8.6 x 10 7 kg CaCO -1