CHAPTER 6 THERMOCHEMISTRY. Questions

CHAPTER 6 THERMOCHEMISTRY Questins 11. Path-dependent functins fr a trip frm Chicag t Denver are thse quantities that depend n the rute taken. One can fly directly frm Chicag t Denver, r ne culd fly frm Chicag t Atlanta t Ls Angeles and then t Denver. Sme path-dependent quantities are miles traveled, fuel cnsumptin f the airplane, time traveling, airplane snacks eaten, etc. State functins are path-independent; they nly depend n the initial and final states. Sme state functins fr an airplane trip frm Chicag t Denver wuld be lngitude change, latitude change, elevatin change, and verall time zne change. 1. Prducts have a lwer ptential energy than reactants when the bnds in the prducts are strnger (n average) than in the reactants. This ccurs generally in exthermic prcesses. Prducts have a higher ptential energy than reactants when the reactants have the strnger bnds (n average). This is typified by endthermic reactins. 1. C 8 H 18 (l) + 5 O (g) 16 CO (g) + 18 H O(g); all cmbustin reactins are exthermic; they all release heat t the surrundings, s q is negative. T determine the sign f w, cncentrate n the mles f gaseus reactants versus the mles f gaseus prducts. In this cmbustin reactin, we g frm 5 mles f reactant gas mlecules t 16 + 18 = 4 mles f prduct gas mlecules. As reactants are cnverted t prducts, an expansin will ccur. When a gas expands, the system des wrk n the surrundings, and w is negative. 14. H = E + P V at cnstant P; frm the definitin f enthalpy, the difference between H and E at cnstant P is the quantity P V. Thus, when a system at cnstant P can d pressurevlume wrk, then H E. When the system cannt d PV wrk, then H = E at cnstant pressure. An imprtant way t differentiate H frm E is t cncentrate n q, the heat flw; the heat flw by a system at cnstant pressure equals H, and the heat flw by a system at cnstant vlume equals E. 15. a. The H value fr a reactin is specific t the cefficients in the balanced equatin. Because the cefficient in frnt f H O is a, 891 kj f heat is released when ml f H O is prduced. Fr 1 ml f H O frmed, 891/ = 446 kj f heat is released. b. 891/ = 446 kj f heat released fr each ml f O reacted. 16. Use the cefficients in the balanced rectin t determine the heat required fr the varius quantities. 90.7 kj a. 1 ml Hg = 90.7 kj required ml Hg 18

CHAPTER 6 THERMOCHEMISTRY 18 b. 1 ml O 90.7 kj 1/ ml O = 181.4 kj required c. When an equatin is reversed, H new = H ld. When an equatin is multiplied by sme integer n, then H new = n( H ld ). Hg(l) + 1/ O (g) HgO(s) Hg(l) + O (g) HgO(s) H = 90.7 kj H = ( 90.7 kj) = 181.4 kj 17. CH 4 (g) + O (g) CO (g) + H O(l) H =!891 kj CH 4 (g) + O (g) CO (g) + H O(g) H =!80 kj H O(l) + 1/ CO (g) 1/ CH 4 (g) + O (g) H 1 =!1/(!891 kj) 1/ CH 4 (g) + O (g) 1/ CO (g) + H O(g) H = 1/(!80 kj) H O(l) H O(g) H = H 1 + H = 44 kj The enthalpy f vaprizatin f water is 44 kj/ml. 18. A state functin is a functin whse change depends nly n the initial and final states and nt n hw ne gt frm the initial t the final state. An extensive prperty depends n the amunt f substance. Enthalpy changes fr a reactin are path-independent, but they d depend n the quantity f reactants cnsumed in the reactin. Therefre, enthalpy changes are a state functin and an extensive prperty. 19. The zer pint fr H f values are elements in their standard state. All substances are measured in relatinship t this zer pint. 0. N matter hw insulated yur therms bttle, sme heat will always escape int the surrundings. If the temperature f the therms bttle (the surrundings) is high, less heat initially will escape frm the cffee (the system); this results in yur cffee staying htter fr a lnger perid f time. 1. Fssil fuels cntain carbn; the incmplete cmbustin f fssil fuels prduces CO(g) instead f CO (g). This ccurs when the amunt f xygen reacting is nt sufficient t cnvert all the carbn t CO. Carbn mnxide is a pisnus gas t humans.. Advantages: H burns cleanly (less pllutin) and gives a lt f energy per gram f fuel. Disadvantages: Expensive and gas strage and safety issues Exercises Ptential and Kinetic Energy 1. KE = mv 1 kg m ; cnvert mass and velcity t SI units. 1 J = s

184 CHAPTER 6 THERMOCHEMISTRY Mass = 5.5 z 1lb 16 z 1 kg = 0.149 kg.05 lb Velcity = 1.0 10 h mi 1 h 60 min 1 min 60 s 1760 yd mi 1 m 1.094 yd = 45 m s 1 KE = mv 1 45 m = 0.149 kg s = 150 J 4. KE = 5 1 mv 1 = 5 1 kg.0 10 cm 1 m 1.0 10 g 1000 g =.0 10 s 100 cm J 5. KE = 1 mv 1 1.0 m =.0 kg s 1 = 1.0 J; KE = mv 1.0 m = 1.0 kg s =.0 J The 1.0-kg bject with a velcity f.0 m/s has the greater kinetic energy. 9.81 m 6. Ball A: PE = mgz =.00 kg s 196 kg m 10.0 m = s = 196 J At pint I: All this energy is transferred t ball B. All f B's energy is kinetic energy at this pint. E ttal = KE = 196 J. At pint II, the sum f the ttal energy will equal 196 J. 9.81 m At pint II: PE = mgz = 4.00 kg s.00 m = 118 J KE = E ttal PE = 196 J 118 J = 78 J Heat and Wrk 7. E = q + w = 45 kj + ( 9 kj) = 16 kj 8. E = q + w = 15 + 104 = 1 kj 9 a. E = q + w =!47 kj + 88 kj = 41 kj b. E = 8! 47 = 5 kj c. E = 47 + 0 = 47 kj d. When the surrundings d wrk n the system, w > 0. This is the case fr a. 0. Step 1: E 1 = q + w = 7 J + 5 J = 107 J; step : E = 5 J 7 J =!7 J E verall = E 1 + E = 107 J 7 J = 70. J

CHAPTER 6 THERMOCHEMISTRY 185 1. E = q + w; wrk is dne by the system n the surrundings in a gas expansin; w is negative. 00. J = q 75 J, q = 75 J f heat transferred t the system. a. E = q + w =! J + 100. J = 77 J b. w =!P V =!1.90 atm(.80 L! 8.0 L) = 10.5 L atm E = q + w = 50. J + 1060 = 1410 J 101. J L atm = 1060 J c. w =!P V =!1.00 atm(9.1 L! 11. L) =!17.9 L atm E = q + w = 107 J! 1810 J =!770 J 101. J L atm =!1810 J. w =!P V; we need the final vlume f the gas. Because T and n are cnstant, P 1 V 1 = P V. V V1 P1 10.0 L(15.0 atm) = = 75.0 L P.00 atm = w =!P V =!.00 atm(75.0 L! 10.0 L) =!10. L atm 4. w =!10. J =!P V,!10 J =!P(5 L! 10. L), P = 14 atm 5. In this prblem, q = w =!950. J. 101. J L atm 1 kj 1000 J =!1. kj = wrk!950. J 1 L atm 101. J w =!P V,!9.8 L atm = =!9.8 L atm f wrk dne by the gases 650. 760 6. E = q + w,!10.5 J = 5.5 J + w, w =!155.0 J atm (V f! 0.040 L), V f! 0.040 = 11.0 L, V f = 11.0 L 1 L atm 101. J w =!P V,!1.50 L atm =!0.500 atm V, V =.06 L V = V f V i,.06 L = 58.0 L! V i, V i = 54.9 L = initial vlume =!1.50 L atm 0.8 J 7. q = mlar heat capacity ml T = 9.1 ml (8.0! 0.0) C = 0,900 J C ml = 0.9 kj 101. J w =!P V =!1.00 atm (998 L! 876 L) =!1 L atm =!1,400 J =!1.4 kj L atm E = q + w = 0.9 kj + (!1.4 kj) = 18.5 kj

186 CHAPTER 6 THERMOCHEMISTRY 8. H O(g) H O(l); E = q + w; q =!40.66 kj; w =!P V Vlume f 1 ml H O(l) = 1.000 ml H O(l) 18.0 g ml 1cm = 18.1 cm = 18.1 ml 0.996 g w =!P V =!1.00 atm (0.0181 L! 0.6 L) = 0.6 L atm E = q + w =!40.66 kj +.10 kj =!7.56 kj 101. J L atm =.10 10 J =.10 kj Prperties f Enthalpy 9. This is an endthermic reactin, s heat must be absrbed in rder t cnvert reactants int prducts. The high-temperature envirnment f internal cmbustin engines prvides the heat. 40. One shuld try t cl the reactin mixture r prvide sme means f remving heat because the reactin is very exthermic (heat is released). The H SO 4 (aq) will get very ht and pssibly bil unless cling is prvided. 41. a. Heat is absrbed frm the water (it gets clder) as KBr disslves, s this is an endthermic prcess. b. Heat is released as CH 4 is burned, s this is an exthermic prcess. c. Heat is released t the water (it gets ht) as H SO 4 is added, s this is an exthermic prcess. d. Heat must be added (absrbed) t bil water, s this is an endthermic prcess. 4. a. The cmbustin f gasline releases heat, s this is an exthermic prcess. b. H O(g) H O(l); heat is released when water vaper cndenses, s this is an exthermic prcess. c. T cnvert a slid t a gas, heat must be absrbed, s this is an endthermic prcess. d. Heat must be added (absrbed) in rder t break a bnd, s this is an endthermic prcess. 4. 4 Fe(s) + O (g) Fe O (s) H = -165 kj; nte that 165 kj f heat is released when 4 ml Fe reacts with ml O t prduce ml Fe O. a. 4.00 ml Fe 165 kj 4 ml Fe =!1650 kj; 1650 kj f heat released b. 1.00 ml Fe O 165 kj ml Fe O =!86 kj; 86 kj f heat released

CHAPTER 6 THERMOCHEMISTRY 187 c. 1.00 g Fe 1 ml Fe 165 kj =!7.9 kj; 7.9 kj f heat released 55.85 g 4 ml Fe d. 10.0 g Fe 1 ml Fe 55.85 g = 0.179 ml Fe;.00 g O 1 ml O.00 g = 0.065 ml O 0.179 ml Fe/0.065 ml O =.86; the balanced equatin requires a 4 ml Fe/ ml O = 1. mle rati. O is limiting since the actual mle Fe/mle O rati is greater than the required mle rati. 0.065 ml O 165 kj ml O 57 kj 44. a. 1.00 ml H O ml H O =!4.4 kj; 4.4 kj f heat released =!86 kj; 86 kj f heat released b. 4.0 g H c. 186 g O 1 ml H 57 kj =!57 kj; 57 kj f heat released.016 g H ml H 1 ml O 57 kj =!0 kj; 0 kj f heat released.00 g O ml O d. n H PV = = RT 1.0 atm.0 10 L 0.0806 L atm 98 K K ml 8 = 8. 10 6 ml H 8. 10 6 57 kj ml H ml H O =!. 10 9 kj;. 10 9 kj f heat released 45. Frm Example 6., q = 1. 10 8 J. Because the heat transfer prcess is nly 60.% efficient, the ttal energy required is 1. 10 8 100. J J =. 10 8 J. 60. J Mass C H 8 =. 10 8 J 1 ml C H 8 1 10 J 44.09 g CH ml C H 8 8 = 4.4 10 g C H 8 46. a. 1.00 g CH 4 1 ml CH4 891 kj =!55.5 kj 16.04 g CH ml CH 4 4 b. n = PV = RT 740. atm 1.00 10 L 760 0.0806 L atm 98 K K ml = 9.8 ml CH 4 9.8 ml 891 kj ml =!.55 10 4 kj

188 CHAPTER 6 THERMOCHEMISTRY 47. When a liquid is cnverted int gas, there is an increase in vlume. The.5 kj/ml quantity is the wrk dne by the vaprizatin prcess in pushing back the atmsphere. 48. H = E + P V; frm this equatin, H > E when V > 0, H < E when V < 0, and H = E when V = 0. Cncentrate n the mles f gaseus prducts versus the mles f gaseus reactants t predict V fr a reactin. a. There are mles f gaseus reactants cnverting t mles f gaseus prducts, s V = 0. Fr this reactin, H = E. b. There are 4 mles f gaseus reactants cnverting t mles f gaseus prducts, s V < 0 and H < E. c. There are 9 mles f gaseus reactants cnverting t 10 mles f gaseus prducts, s V > 0 and H > E. Calrimetry and Heat Capacity 49. Specific heat capacity is defined as the amunt f heat necessary t raise the temperature f 1 gram f substance by 1 degree Celsius. Therefre, H O(l) with the largest heat capacity value requires the largest amunt f heat fr this prcess. The amunt f heat fr H O(l) is: energy = s m T = 5.0 g (7.0 C!15.0 C) =.0 10 J The largest temperature change when a certain amunt f energy is added t a certain mass f substance will ccur fr the substance with the smallest specific heat capacity. This is Hg(l), and the temperature change fr this prcess is: T = energy s m = 10.7 kj 0.14 J 1000 J kj 550. g = 140 C 50. a. s = specific heat capacity = 0.4 J 0.4 J = since T(K) = T( C). K g Energy = s m T = 0.4 J 150.0 g (98 K - 7 K) = 9.0 10 J b. Mlar heat capacity = 0.4 J 107.9 g Ag ml Ag = 6 J C ml c. 150 J = 0.4 J m (15. C! 1.0 C), m = 150 = 1.6 10 g Ag 0.4.

CHAPTER 6 THERMOCHEMISTRY 189 51. s = specific heat capacity = q m T 1 J = = 0.890 J/ECCg 5.00 g (55.1 5.) C Frm Table 6.1, the substance is slid aluminum. 5. s = 585 J 15.6 g (5.5 0.0) C = 0.19 J/ C Mlar heat capacity = 0.19 J 00.6 g mlhg = 7.9 J C ml 5. Heat lss by ht water = heat gain by cler water The magnitudes f heat lss and heat gain are equal in calrimetry prblems. The nly difference is the sign (psitive r negative). T avid sign errrs, keep all quantities psitive and, if necessary, deduce the crrect signs at the end f the prblem. Water has a specific heat capacity = s = / CCg = /KCg ( T in C = T in K). Heat lss by ht water = s m T = Heat gain by cler water = 09 J 15 J (0. K Tf ) = (Tf 80. K) K K 50.0 g (0. K! T f ) K g 0.0 g (T f!80. K); heat lss = heat gain, s: K g 6.90 10 4! 09T f = 15T f.50 10 4, 4T f = 1.040 10 5, T f = 11 K Nte that the final temperature is clser t the temperature f the mre massive ht water, which is as it shuld be. 54. Heat gained by water = heat lst by nickel = s m T, where s = specific heat capacity. Heat gain = 4.18J 150.0 g (5.0 C.5 C) = 940 J A cmmn errr in calrimetry prblems is sign errrs. Keeping all quantities psitive helps t eliminate sign errrs. Heat lss = 940 J = 0.444 J mass (99.8 5.0) C, mass = 940 0.444 74.8 = 8 g 55. Heat lss by Al + heat lss by Fe = heat gain by water; keeping all quantities psitive t avid sign errr:

190 CHAPTER 6 THERMOCHEMISTRY 0.89 J 5.00 g Al (100.0 C T f ) + 0.45 J 10.00 g Fe (100.0 T f ) = 97. g H O (T f.0 C) 4.5(100.0 T f ) + 4.5(100.0 T f ) = 407(T f.0), 450 (4.5)T f + 450 (4.5)T f = 407T f 8950 416T f = 9850, T f =.7 C 10. J 56. Heat released t water = 5.0 g H g H + 10. g methane 50. J g methane = 1.10 10 J Heat gain by water = 1.10 10 J = 50.0 g T T = 5.6 C, 5.6 C = T f! 5.0 C, T f = 0. C 57. Heat gain by water = heat lss by metal = s m T, where s = specific heat capacity. Heat gain = 150.0 g (18. C - 15.0 C) = 100 J A cmmn errr in calrimetry prblems is sign errrs. Keeping all quantities psitive helps t eliminate sign errrs. Heat lss = 100 J = s 150.0 g (75.0 C - 18. C), s = 100 J 150.0 g 56.7 = 0.5 J/ CCg C 58. Heat gain by water = heat lss by Cu; keeping all quantities psitive helps t avid sign errrs: mass (4.9 C!. C) = 0.0 J 110. g Cu (8.4 C! 4.9 C) 11(mass) = 100, mass = 10 g H O 59. 50.0 10 L 0.100 ml/l = 5.00 10 ml f bth AgNO and HCl are reacted. Thus 5.00 10 ml f AgCl will be prduced because there is a 1 : 1 mle rati between reactants. Heat lst by chemicals = heat gained by slutin Heat gain = 100.0 g (.40.60) C = 0 J

CHAPTER 6 THERMOCHEMISTRY 191 Heat lss = 0 J; this is the heat evlved (exthermic reactin) when 5.00 10 ml f AgCl is prduced. S q = 0 J and H (heat per ml AgCl frmed) is negative with a value f: H = 0 J 5.00 10 ml 1 kj 1000 J = 66 kj/ml Nte: Sign errrs are cmmn with calrimetry prblems. Hwever, the crrect sign fr H can be determined easily frm the T data; i.e., if T f the slutin increases, then the reactin is exthermic because heat was released, and if T f the slutin decreases, then the reactin is endthermic because the reactin absrbed heat frm the water. Fr calrimetry prblems, keep all quantities psitive until the end f the calculatin and then decide the sign fr H. This will help t eliminate errrs. 60. NaOH(aq) + HCl(aq) NaCl(aq) + H O(l) We have a stichimetric mixture. All f the NaOH and HCl will react. 0.10 L 1.0 ml = 0.10 ml f HCl is neutralized by 0.10 ml NaOH. L Heat lst by chemicals = heat gained by slutin Vlume f slutin = 100.0 + 100.0 = 00.0 ml Heat gain = 4.18J 1.0 g 00.0 ml ml (1. 4.6) C = 5.6 10 J = 5.6 kj Heat lss = 5.6 kj; this is the heat released by the neutralizatin f 0.10 ml HCl. Because the temperature increased, the sign fr H must be negative, i.e., the reactin is exthermic. Fr calrimetry prblems, keep all quantities psitive until the end f the calculatin and then decide the sign fr H. H = 5.6 kj 0.10 ml = 56 kj/ml 61. Heat lst by slutin = heat gained by KBr; mass f slutin = 15 g + 10.5 g = 16 g Nte: Sign errrs are cmmn with calrimetry prblems. Hwever, the crrect sign fr H can easily be btained frm the T data. When wrking calrimetry prblems, keep all quantities psitive (ignre signs). When finished, deduce the crrect sign fr H. Fr this prblem, T decreases as KBr disslves, s H is psitive; the disslutin f KBr is endthermic (absrbs heat). Heat lst by slutin = 16 g (4. C 1.1 C) = 1800 J = heat gained by KBr

19 CHAPTER 6 THERMOCHEMISTRY H in units f J/g = 1800 J 10.5 g KBr = 170 J/g H in units f kj/ml = 170 J g KBr 119.0 g KBr 1 kj = 0. kj/ml ml KBr 1000 J 6. NH 4 NO (s) NH 4 + (aq) + NO (aq) H =?; mass f slutin = 75.0 g + 1.60 g = 76.6 g Heat lst by slutin = heat gained as NH 4 NO disslves. T help eliminate sign errrs, we will keep all quantities psitive (q and T) and then deduce the crrect sign fr H at the end f the prblem. Here, because temperature decreases as NH 4 NO disslves, heat is absrbed as NH 4 NO disslves, s this is an endthermic prcess ( H is psitive). Heat lst by slutin = 76.6 g (5.00.4) C = 5 J = heat gained as NH 4 NO disslves H = 5 J 1.60 g NH NO 4 80.05 g NH4NO 1 kj = 6.6 kj/ml NH 4 NO disslving ml NH NO 1000 J 4 6. Because H is exthermic, the temperature f the slutin will increase as CaCl (s) disslves. Keeping all quantities psitive: 1 ml CaCl 81.5 kj heat lss as CaCl disslves = 11.0 g CaCl = 8.08 kj 110.98 g CaCl ml CaCl heat gained by slutin = 8.08 10 J = T f 5.0 C = 8.08 4.18 (15 + 11.0) g (T f 5.0 C) 10 = 14. C, T f = 14. C + 5.0 C = 9. C 16 0.500 ml HCl 64. 0.100 L L = 5.00 10 ml HCl 0.100 ml Ba(OH) 0.00 L =.00 10 ml Ba(OH) L T react with all the HCl present, 5.00 10 / =.50 10 ml Ba(OH) is required. Because 0.000 ml Ba(OH) is present, HCl is the limiting reactant. 5.00 10 ml HCl 118 kj ml HCl Heat gained by slutin =.95 10 J = T = 1.76 C = T f T i = T f 5.0 C, T f = 6.8 C =.95 kj f heat is evlved by reactin 400.0 g T

CHAPTER 6 THERMOCHEMISTRY 19 65. a. Heat gain by calrimeter = heat lss by CH 4 = 6.79 g CH 4 1 ml CH 80 kj 16.04 g ml = 40. kj 40. kj Heat capacity f calrimeter = = 1.5 kj/ C 10.8 C 1. 5 kj b. Heat lss by C H = heat gain by calrimeter = 16.9 C = 5 kj C 5 kj 6.04 g E cmb = 1.6 g C H ml C =!1.10 10 kj/ml H Nte: Because bmb calrimeters are at cnstant vlume, q V = E. 66. First, we need t get the heat capacity f the calrimeter frm the cmbustin f benzic acid. Heat lst by cmbustin = heat gained by calrimeter. 6.4 kj Heat lss = 0.1584 g = 4.185 kj g 4.185 kj Heat gain = 4.185 kj = C cal T, C cal = = 1.65 kj/ C.54 C Nw we can calculate the heat f cmbustin f vanillin. Heat lss = heat gain. 1.65 kj Heat gain by calrimeter =.5 C = 5.6 kj C Heat lss = 5.6 kj, which is the heat evlved by cmbustin f the vanillin. E cmb = Hess's Law 5.6 kj 0.10 g = 5. kj/g; E cmb = 5. kj g 15.14 g ml = 80 kj/ml 67. Infrmatin given: C(s) + O (g) CO (g) CO(g) + 1/ O (g) CO (g) H =!9.7 kj H =!8. kj Using Hess s law: C(s) + O (g) CO (g) H 1 = (!9.7 kj) CO (g) CO(g) + O (g) H =!(!8. kj) C(s) + O (g) CO(g) H = H 1 + H =!0.8 kj Nte: The enthalpy change fr a reactin that is reversed is the negative quantity f the enthalpy change fr the riginal reactin. If the cefficients in a balanced reactin are multiplied by an integer, then the value f H is multiplied by the same integer.

194 CHAPTER 6 THERMOCHEMISTRY 68. C 4 H 4 (g) + 5 O (g) 4 CO (g) + H O(l) H cmb =!41 kj C 4 H 8 (g) + 6 O (g) 4 CO (g) + 4 H O(l) H cmb =!755 kj H (g) + 1/ O (g) H O(l) H cmb =!86 kj By cnventin, H O(l) is prduced when enthalpies f cmbustin are given, and because per-mle quantities are given, the cmbustin reactin refers t 1 mle f that quantity reacting with O (g). Using Hess s law t slve: C 4 H 4 (g) + 5 O (g) 4 CO (g) + H O(l) H 1 =!41 kj 4 CO (g) + 4 H O(l) C 4 H 8 (g) + 6 O (g) H =! (!755 kj) H (g) + O (g) H O(l) H = (!86 kj) C 4 H 4 (g) + H (g) C 4 H 8 (g) H = H 1 + H + H =!158 kj 69. N (g) + 6 H (g) 4 NH (g) H =!4(46 kj) 6 H O(g) 6 H (g) + O (g) H =!(-484 kj) N (g) + 6 H O(g) O (g) + 4 NH (g) H = 168 kj N, because the reactin is very endthermic (requires a lt f heat t react), it wuld nt be a practical way f making ammnia because f the high energy csts required. 70. ClF + 1/ O 1/ Cl O + 1/ F O H = 1/(167.4 kj) 1/ Cl O + / F O ClF + O H =!1/(41.4 kj) F + 1/ O F O H = 1/(!4.4 kj) ClF(g) + F (g) ClF H =!108.7 kj 71. NO + O NO + O H =!199 kj / O O H =!1/(-47 kj) O 1/ O H =!1/(495 kj) NO(g) + O(g) NO (g) H =! kj 7. We want H fr N H 4 (l) + O (g) N (g) + H O(l). It will be easier t calculate H fr the cmbustin f fur mles f N H 4 because we will avid fractins. 9 H + 9/ O 9 H O H = 9( 86 kj) N H 4 + H O N O + 9 H H = ( 17 kj) NH + N O 4 N + H O H = 1010. kj N H 4 + H O NH + 1/ O H = ( 14 kj) 4 N H 4 (l) + 4 O (g) 4 N (g) + 8 H O(l) H = 490. kj 490. kj Fr N H 4 (l) + O (g) N (g) + H O(l) H = 4 = 6 kj

CHAPTER 6 THERMOCHEMISTRY 195 Nte: By the significant figure rules, we culd reprt this answer t fur significant figures. Hwever, because the H values given in the prblem are nly knwn t ±1 kj, ur final answer will at best be ±1 kj. 7. CaC Ca + C H =!(!6.8 kj) CaO + H O Ca(OH) H =!65.1 kj CO + H O C H + 5/ O H =!(!100. kj) Ca + 1/ O CaO H =!65.5 kj C + O CO H = (!9.5 kj) CaC (s) + H O(l) Ca(OH) (aq) + C H (g) H =!71 kj 74. P 4 O 10 P 4 + 5 O H =!(!967. kj) 10 PCl + 5 O 10 Cl PO H = 10(!85.7 kj) 6 PCl 5 6 PCl + 6 Cl H =!6(!84. kj) P 4 + 6 Cl 4 PCl H =!15.6 P 4 O 10 (s) + 6 PCl 5 (g) 10 Cl PO(g) H =!610.1 kj Standard Enthalpies f Frmatin 75. The change in enthalpy that accmpanies the frmatin f 1 mle f a cmpund frm its elements, with all substances in their standard states, is the standard enthalpy f frmatin fr a cmpund. The reactins that refer t H f are: Na(s) + 1/ Cl (g) NaCl(s); H (g) + 1/ O (g) H O(l) 6 C(graphite, s) + 6 H (g) + O (g) C 6 H 1 O 6 (s) Pb(s) + S(rhmbic, s) + O (g) PbSO 4 (s) 76. a. Aluminum xide = Al O ; Al(s) + / O (g) Al O (s) b. C H 5 OH(l) + O (g) CO (g) + H O(l) c. NaOH(aq) + HCl(aq) H O(l) + NaCl(aq) d. C(graphite, s) + / H (g) + 1/ Cl (g) C H Cl(g) e. C 6 H 6 (l) + 15/ O (g) 6 CO (g) + H O(l) Nte: H cmb values assume 1 mle f cmpund cmbusted. f. NH 4 Br(s) NH 4 + (aq) + Br - (aq) 77. In general, H = n p H f! n, prducts r H f, and all elements in their standard, reactants state have = 0 by definitin. H f a. The balanced equatin is NH (g) + O (g) + CH 4 (g) HCN(g) + 6 H O(g).

196 CHAPTER 6 THERMOCHEMISTRY H = ( ml HCN H f, HCN + 6 ml H O(g) H f, HO )! ( ml NH + ml CH 4 H ) H f, NH f, CH 4 H = [(15.1) + 6(!4)]! [(!46) + (!75)] =!940. kj b. Ca (PO 4 ) (s) + H SO 4 (l) CaSO 4 (s) + H PO 4 (l) H = 14 kj (s) + ml H ml ml CaSO4 4 167 kj PO (l) ml 416 kj 814 kj! 1 ml Ca (PO4) (s) + ml HSO4(l) ml ml H =!68 kj! (!6568 kj) =!65 kj c. NH (g) + HCl(g) NH 4 Cl(s) H = (1 ml NH 4 Cl )! (1 ml NH H f, NH4Cl H f, + 1 ml HCl H f, HCl NH ) 14 kj 46 kj 9 kj H = 1 ml 1 ml + 1 ml ml ml ml H =!14 kj + 18 kj =!176 kj 78. a. The balanced equatin is C H 5 OH(l) + O (g) CO (g) + H O(g). H = 9.5 kj ml ml + 4 kj ml ml H =!151 kj! (!78 kj) =!15 kj b. SiCl 4 (l) + H O(l) SiO (s) + 4 HCl(aq) 78 kj 1 ml ml Because HCl(aq) is H + (aq) + Cl (aq), H f = 0! 167 =!167 kj/ml. H = 167 kj 4 ml ml + 911 kj 1 ml ml 687 kj 86 kj 1 ml + ml ml ml H =!1579 kj! (!159 kj) =!0. kj c. MgO(s) + H O(l) Mg(OH) (s) H = 95 kj 1 ml ml 60 kj 86 kj 1 ml + 1 ml ml ml H =!95 kj! (!888 kj) =!7 kj

CHAPTER 6 THERMOCHEMISTRY 197 79. a. 4 NH (g) + 5 O (g) 4 NO(g) + 6 H O(g); H = n p H f! n, prducts r H f, reactants H = 90. kj 4 kj 46 kj 4 ml + 6 ml 4 ml =!908 kj ml ml ml NO(g) + O (g) NO (g) H = 4 kj 90. kj ml ml =!11 kj ml ml NO (g) + H O(l) HNO (aq) + NO(g) H = Nte: All 07 kj 90. kj ml + 1 ml ml ml H f values are assumed ±1 kj. 4 kj 86 kj ml + 1 ml ml ml =!140. kj b. 1 NH (g) + 15 O (g) 1 NO(g) + 18 H O(g) 1 NO(g) + 6 O (g) 1 NO (g) 1 NO (g) + 4 H O(l) 8 HNO (aq) + 4 NO(g) 4 H O(g) 4 H O(l) 1 NH (g) + 1 O (g) 8 HNO (aq) + 4 NO(g) + 14 H O(g) The verall reactin is exthermic because each step is exthermic. 416 kj 80. 4 Na(s) + O (g) Na O(s) H = ml ml = 8 kj Na(s) + H O(l) NaOH(aq) + H (g) H = 470. kj 86 kj ml ml = 68 kj ml ml Na(s) + CO (g) Na O(s) + CO(g) 416 kj 110.5 kj 9.5 kj H = 1 ml + 1 ml 1 ml = 1 kj ml ml ml In Reactins and, sdium metal reacts with the "extinguishing agent." Bth reactins are exthermic, and each reactin prduces a flammable gas, H and CO, respectively. 81. Al(s) + NH 4 ClO 4 (s) Al O (s) + AlCl (s) + NO(g) + 6 H O(g)

198 CHAPTER 6 THERMOCHEMISTRY H = 4 kj 90. kj 6 ml + ml ml ml + 704 kj 1676 kj 1 ml + 1 ml ml ml 95 kj ml =!677 kj ml 8. 5 N O 4 (l) + 4 N H CH (l) 1 H O(g) + 9 N (g) + 4 CO (g) H = 4 kj 9.5 kj 1 ml + 4 ml ml ml 0. kj 54 kj 5 ml + 4 ml =!4594 kj ml ml 8. ClF (g) + NH (g) N (g) + 6 HF(g) + Cl (g) H = 1196 kj H f, HF, ClF f, NH H = (6 ) ( H + H ) f 71 kj 1196 kj = 6 ml ml 46 kj H f, ClF ml ml H f, ClF 1196 kj = 166 kj + 9 kj, H f, ClF = ( 166 + 9 + 1196) kj = ml 169 kj ml 84. C H 4 (g) + O (g) CO (g) + H O(l) H =!1411.1 kj H =!1411.1 kj = (!9.5) kj + (!85.8) kj! f, C H 4 H f, C H 4 H f, C H 4!1411.1 kj =!158.6 kj! H, = 5.5 kj/ml Energy Cnsumptin and Surces 85. C(s) + H O(g) H (g) + CO(g) H = 110.5 kj ( 4 kj) = 1 kj 86. CO(g) + H (g) CH OH(l) H = 9 kj ( 110.5 kj) = 19 kj 87. C H 5 OH(l) + O (g) CO (g) + H O(l) H = [(!9.5 kj) + (!86 kj)]! (!78 kj) =!167 kj/ml ethanl 167 kj ml 1 ml 46.07 g =!9.67 kj/g 88. CH OH(l) + / O (g) CO (g) + H O(l) H = [!9.5 kj + (!86 kj)]! (!9 kj) =!77 kj/ml CH OH

CHAPTER 6 THERMOCHEMISTRY 199 77 kj ml 1 ml.04 g =!.7 kj/g versus!9.67 kj/g fr ethanl Ethanl has a slightly higher fuel value than methanl. 89. C H 8 (g) + 5 O (g) CO (g) + 4 H O(l) H = [(!9.5 kj) + 4(!86 kj)]! (!104 kj) =!1 kj/ml C H 8 1 kj ml 1 ml 44.09 g = 50.7 kj ml versus!47.7 kj/g fr ctane (Example 6.11) The fuel values are very clse. An advantage f prpane is that it burns mre cleanly. The biling pint f prpane is!4 C. Thus it is mre difficult t stre prpane, and there are extra safety hazards assciated with using high-pressure cmpressed-gas tanks. 90. 1 mle f C H (g) and 1 mle f C 4 H 10 (g) have equivalent vlumes at the same T and P. Enthalpy f cmbustin per vlume f CH Enthalpy f cmbustin per vlume f C H 4 10 = enthalpy f cmbustin per ml f enthalpy f cmbustin per ml f C C 4 H H 10 Enthalpy f cmbustin per vlume f CH Enthalpy f cmbustin per vlume f C H 4 10 = 49.9 kj 6.04 g CH g CH ml CH 49.5 kj 58.1 g C4H g C H ml C H 4 10 4 10 10 = 0.45 Mre than twice the vlume f acetylene is needed t furnish the same energy as a given vlume f butane. 91. The mlar vlume f a gas at STP is.4 L (frm Chapter 5). 4.19 10 6 1 ml CH kj 891 kj 4.4 L CH4 = 1.05 10 5 L CH 4 ml CH 4 9. Mass f H O = 1.00 gal.785 L gal Energy required (theretical) = s m T = 1000 ml 1.00 g = 790 g H O L ml 790 g 10.0 C = 1.58 10 5 J Fr an actual (80.0% efficient) prcess, mre than this quantity f energy is needed since heat is always lst in any transfer f energy. The energy required is: 1.58 10 5 100. J J = 1.98 10 5 J 80.0 J Mass f C H = 1.98 10 5 1 ml CH J 100. 10 J 6.04 g C ml C H H =.97 g C H

00 CHAPTER 6 THERMOCHEMISTRY Cnnecting t Bichemistry 9. Because the sign f H is negative, the reactin is exthermic. Heat is evlved by the system t the surrundings. 94. Frm the prblem, walking 4.0 miles cnsumes 400 kcal f energy. 454 g 7.7 kcal 4 mi 1 h 1 lb fat = 8.7 h = 9 h lb g 400 kcal 4 mi 95..0 h 5500 kj 1 ml H O 18.0 g HO = 4900 g = 4.9 kg H O h 40.6 kj ml 96. Heat lss by ht water = heat gain by cld water; keeping all quantities psitive helps t avid sign errrs: m ht (55.0 C! 7.0 C) = 90.0 g (7.0 C!.0 C) m ht = 90.0 g 15.0 18.0 C C = 75.0 g ht water needed 1.56 kj 97. Heat gain by calrimeter =. C = 5.0 kj = heat lss by quinine C Heat lss = 5.0 kj, which is the heat evlved (exthermic reactin) by the cmbustin f 0.1964 g f quinne. Because we are at cnstant vlume, q V = E. E cmb = 5.0 kj 0.1964 g 5 kj =!5 kj/g; E cmb = g 98. a. C 1 H O 11 (s) + 1 O (g) 1 CO (g) + 11 H O(l) 108.09 g ml b. A bmb calrimeter is at cnstant vlume, s heat released = q V = E: 4.00 kj E = 1.46 g 4.0 g ml =!560 kj/ml C 1 H O 11 =!700 kj/ml c. PV = nrt; at cnstant P and T, P V = RT n, where n = mles f gaseus prducts! mles f gaseus reactants. H = E + P V = E + RT n Fr this reactin, n = 1!1 = 0, s H = E =!560 kj/ml. 99. C 6 H 4 (OH) C 6 H 4 O + H H = 177.4 kj H O H + O H =! (!191. kj) H + O H O(g) H = (!41.8 kj) H O(g) H O(l) H = (!4.8 kj) C 6 H 4 (OH) (aq) + H O (aq) C 6 H 4 O (aq) + H O(l) H =!0.6 kj

CHAPTER 6 THERMOCHEMISTRY 01 100. Frm the phtsynthesis reactin, CO (g) is used by plants t cnvert water int glucse and xygen. If the plant ppulatin is significantly reduced, nt as much CO will be cnsumed in the phtsynthesis reactin. As the CO levels f the atmsphere increase, the greenhuse effect due t excess CO in the atmsphere will becme wrse. Additinal Exercises 101. a. SO (g) + O (g) SO (g); w = P V; because the vlume f the pistn apparatus decreased as reactants were cnverted t prducts ( V < 0), w is psitive (w > 0). b. COCl (g) CO(g) + Cl (g); because the vlume increased ( V > 0), w is negative (w < 0). c. N (g) + O (g) NO(g); because the vlume did nt change ( V = 0), n PV wrk is dne (w = 0). In rder t predict the sign f w fr a reactin, cmpare the cefficients f all the prduct gases in the balanced equatin t the cefficients f all the reactant gases. When a balanced reactin has mre mles f prduct gases than mles f reactant gases (as in b), the reactin will expand in vlume ( V psitive), and the system des wrk n the surrundings. When a balanced reactin has a decrease in the mles f gas frm reactants t prducts (as in a), the reactin will cntract in vlume ( V negative), and the surrundings will d cmpressin wrk n the system. When there is n change in the mles f gas frm reactants t prducts (as in c), V = 0 and w = 0. 10. w = P V; n = mles f gaseus prducts mles f gaseus reactants. Only gases can d PV wrk (we ignre slids and liquids). When a balanced reactin has mre mles f prduct gases than mles f reactant gases ( n psitive), the reactin will expand in vlume ( V psitive), and the system will d wrk n the surrundings. Fr example, in reactin c, n = 0 = mles, and this reactin wuld d expansin wrk against the surrundings. When a balanced reactin has a decrease in the mles f gas frm reactants t prducts ( n negative), the reactin will cntract in vlume ( V negative), and the surrundings will d cmpressin wrk n the system, e.g., reactin a, where n = 0 1 = 1. When there is n change in the mles f gas frm reactants t prducts, V = 0 and w = 0, e.g., reactin b, where n = = 0. When V > 0 ( n > 0), then w < 0, and the system des wrk n the surrundings (c and e). When V < 0 ( n < 0), then w > 0, and the surrundings d wrk n the system (a and d). When V = 0 ( n = 0), then w = 0 (b). 10. E verall = E step 1 + E step ; this is a cyclic prcess, which means that the verall initial state and final state are the same. Because E is a state functin, E verall = 0 and E step 1 =! E step.

0 CHAPTER 6 THERMOCHEMISTRY E step 1 = q + w = 45 J + (!10. J) = 5 J E step =! E step 1 =!5 J = q + w,!5 J =!60 J + w, w = 5 J 104. K(s) + H O(l) KOH(aq) + H (g) H = (!481 kj)! (!86 kj) =!90. kj 1 ml K 90. kj 5.00 g K =!4.9 kj 9.10 g K ml K 4.9 kj f heat is released n reactin f 5.00 g K. 4,900 J = (1.00 10 4,900 g) T, T = = 5.96 C g C 4.18 1.00 10 Final temperature = 4.0 + 5.96 = 0.0 C 105. HCl(aq) + NaOH(aq) H O(l) + NaCl(aq) H =!56 kj 0.400 ml HCl 0.000 L L = 8.00 10 ml HCl 0.1500 L 0.500 ml NaOH L = 7.50 10 ml NaOH Because the balanced reactin requires a 1 : 1 mle rati between HCl and NaOH, and because fewer mles f NaOH are actually present as cmpared with HCl, NaOH is the limiting reagent. 7.50 10 ml NaOH 56 kj ml NaOH =!4. kj; 4. kj f heat is released. 106. Na SO 4 (aq) + Ba(NO )(aq) BaSO 4 (s) + NaNO (aq) H =? 1.00 L.00 ml L =.00 ml Na SO 4 ;.00 L 0.750 ml L = 1.50 ml Ba(NO ) The balanced equatin requires a 1 : 1 mle rati between Na SO 4 and Ba(NO ). Because we have fewer mles f Ba(NO ) present, it is limiting, and 1.50 ml BaSO 4 will be prduced [there is a 1 : 1 mle rati between Ba(NO ) and BaSO 4 ]. Heat gain by slutin = heat lss by reactin Mass f slutin =.00 L Heat gain by slutin = 1000 ml 1 L.00 g = 6.00 10 g ml 6.7 J 6.00 10 g (4.0! 0.0)EC = 4.59 10 5 J

CHAPTER 6 THERMOCHEMISTRY 0 Because the slutin gained heat, the reactin is exthermic; q =!4.59 10 5 J fr the reactin. H = 4.59 10 J 1.50 ml BaSO 5 4 =!.06 10 5 J/ml =!06 kj/ml 107. q surr = q slutin + q cal ; we nrmally assume that q cal is zer (n heat gain/lss by the calrimeter). Hwever, if the calrimeter has a nnzer heat capacity, then sme f the heat absrbed by the endthermic reactin came frm the calrimeter. If we ignre q cal, then q surr is t small, giving a calculated H value that is less psitive (smaller) than it shuld be. 108. The specific heat f water is / CCg, which is equal t 4.18 kj/ CCkg. We have 1.00 kg f H O, s: 1.00 kg = 4.18 kj/ C C kg This is the prtin f the heat capacity that can be attributed t H O. Ttal heat capacity = C cal + C H O, C cal = 10.84! 4.18 = 6.66 kj/ C 109. Heat released = 1.056 g 6.4 kj/g = 7.90 kj = heat gain by water and calrimeter Heat gain = 7.90 kj = C kg 6.66 kj 0.987 kg T + T C 7.90 = (4.1 + 6.66) T = (10.79) T, T =.586 C.586 C = T f!. C, T f = 5.91 C 110. Fr Exercise 81, a mixture f ml Al and ml NH 4 ClO 4 yields 677 kj f energy. The mass f the stichimetric reactant mixture is: 6.98 g 117.49 g ml + ml = 4.41 g ml ml Fr 1.000 kg f fuel: 1.000 10 g 677 kj 4.41 g = 6177 kj In Exercise 8, we get 4594 kj f energy frm 5 ml f N O 4 and 4 ml f N H CH. The 9.0 g 46.08 g mass is 5 ml + 4 ml = 644.4 kj. ml ml Fr 1.000 kg f fuel: 1.000 10 g 4594 kj 644.4 g = 719 kj Thus we get mre energy per kilgram frm the N O 4 /N H CH mixture.

04 CHAPTER 6 THERMOCHEMISTRY 111. 1/ D 1/ A + B H = 1/6( 40 kj) 1/ E + F 1/ A H = 1/( 105. kj) 1/ C 1/ E + / D H = 1/(64.8 kj) F + 1/ C A + B + D H = 47.0 kj 11. T avid fractins, let's first calculate H fr the reactin: 6 FeO(s) + 6 CO(g) 6 Fe(s) + 6 CO (g) 6 FeO + CO Fe O 4 + CO H =!(18 kj) Fe O 4 + CO Fe O + CO H =! (!9 kj) Fe O + 9 CO 6 Fe + 9 CO H = (! kj) 6 FeO(s) + 6 CO(g) 6 Fe(s) + 6 CO (g) H =!66 kj S fr FeO(s) + CO(g) Fe(s) + CO (g), H = 11. a. H = ml(7 kj/ml)! 1 ml(49 kj/ml) = 6 kj 66 kj =!11 kj. 6 b. Because C H (g) is higher in energy than C 6 H 6 (l), acetylene will release mre energy per gram when burned in air. 114. I(g) + Cl(g) ICl(g) H =!(11. kj) 1/ Cl (g) Cl(g) H = 1/(4. kj) 1/ I (g) I(g) H = 1/(151.0 kj) 1/ I (s) 1/ I (g) H = 1/(6.8 kj) 1/ I (s) + 1/ Cl (g) ICl(g) H = 16.8 kj/ml = H f, ICl 115. a. C H 4 (g) + O (g) CH CHO(g) + O (g) H =!166 kj! [14 kj + 5 kj] =!61 kj b. O (g) + NO(g) NO (g) + O (g) H = 4 kj! [90. kj + 14 kj] =!199 kj c. SO (g) + H O(l) H SO 4 (aq) H =!909 kj![!96 kj + (!86 kj)] =!7 kj d. NO(g) + O (g) NO (g) H = (4) kj! (90.) kj =!11 kj

CHAPTER 6 THERMOCHEMISTRY 05 Challenge Prblems 116. Only when there is a vlume change can PV wrk be dne. In pathway 1 (steps 1 + ), nly the first step des PV wrk (step has a cnstant vlume f 0.0 L). In pathway (steps + 4), nly step 4 des PV wrk (step has a cnstant vlume f 10.0 L). Pathway 1: w =!P V =!.00 atm(0.0 L! 10.0 L) = -40.0 L atm 101. J L atm =!4.05 10 J Pathway : w =!P V =!1.00 atm(0.0 L! 10.0 L) =!0.0 L atm 101. J L atm =!.0 10 J Nte: The sign is (!) because the system is ding wrk n the surrundings (an expansin). We get different values f wrk fr the tw pathways; bth pathways have the same initial and final states. Because w depends n the pathway, wrk cannt be a state functin. 117. A(l) A(g) H vap = 0.7 kj w = P V = nrt, where n = n prducts n reactants = 1 0 = 1 w = (1 ml)(8.145 J/KCml)(80. + 7 K) = 940 J =.94 kj Because pressure is cnstant: E = q p + w = H + w = 0.7 kj + (.94 kj) = 7.8 kj 118. 0. 10 g C1HO11 1 ml C1HO11 Energy needed = h 4. g C1HO11 Energy frm sun = 1.0 kw/m = 1000 W/m 1000 J 1.0 kj = = s m s m 10,000 m 1.0 kj 60 s 60 min =.6 10 7 kj/h s m min h 5640 kj ml =. 10 5 kj/h Percent efficiency = energy used per hur ttal energy per hur 100 =..6 10 10 5 7 kj 100 = 0.9% kj 40. kj h 119. Energy used in 8.0 hurs = 40. kwh = s 600 s = 1.4 10 5 kj h 10. kj Energy frm the sun in 8.0 hurs = s m 60 s 60 min 8.0 h =.9 10 4 kj/m min h Only 1% f the sunlight is cnverted int electricity: 0.1 (.9 10 4 kj/m ) area = 1.4 10 5 kj, area = 7 m

06 CHAPTER 6 THERMOCHEMISTRY 10. a. HNO (aq) + Na CO (s) NaNO (aq) + H O(l) + CO (g) H = [(!467 kj) + (!86 kj) + (!9.5 kj)]! [(!07 kj) + (!111 kj)] =!69 kj.0 10 4 4 qt 946 ml 1.4 g gallns = 1.1 10 8 g f cncentrated nitric gal qt ml acid slutin 1.1 10 8 70.0 g HNO g slutin = 7.7 10 7 g HNO 100.0 g slutin 7.7 10 7 g HNO 1 ml 6.0 g 1 ml Na CO ml HNO 105.99 g Na CO ml Na CO = 6.5 10 7 g Na CO There are (7.7 10 7 /6.0) ml f HNO frm the previus calculatin. There are 69 kj f heat evlved fr every mles f nitric acid neutralized. Cmbining these tw results: 7.7 10 7 1 ml HNO 69 kj g HNO =!4. 10 7 kj 6.0 g HNO ml HNO b. They feared the heat generated by the neutralizatin reactin wuld vaprize the unreacted nitric acid, causing widespread airbrne cntaminatin. 11. 400 kcal 4.18 kj kcal = 1.7 10 kj. 10 kj PE = mgz = 1 kg 9.81 m.54 cm 1 m 180 lb 8 in = 160 J. 00 J.05 lb s in 100 cm 00 J f energy is needed t climb ne step. The ttal number f steps t climb are: 10 6 J 1step 00 J = 1 10 4 steps 1. H (g) + 1/ O (g) H O(l) H =, H O(l) =!85.8 kj; we want the reverse reactin: H f H O(l) H (g) + 1/ O (g) H = 85.8 kj w =!P V; because PV = nrt, at cnstant T and P, P V = RT n, where n = mles f gaseus prducts mles f gaseus reactants. Here, n = (1 ml H + 0.5 ml O ) (0) = 1.5 ml. E = H!P V = H! nrt E = 85.8 kj! 1.50 ml 8.145 J/K E = 85.8 kj!.7 kj = 8.1 kj ml 98 K 1 kj 1000 J

CHAPTER 6 THERMOCHEMISTRY 07 1. There are five parts t this prblem. We need t calculate: (1) q required t heat H O(s) frm!0.ec t 0EC; use the specific heat capacity f H O(s) () q required t cnvert 1 ml H O(s) at 0EC int 1 ml H O(l) at 0EC; use H fusin () q required t heat H O(l) frm 0EC t 100.EC; use the specific heat capacity f H O(l) (4) q required t cnvert 1 ml H O(l) at 100.EC int 1 ml H O(g) at 100.EC; use H vaprizatin (5) q required t heat H O(g) frm 100.EC t 140.EC; use the specific heat capacity f H O(g) We will sum up the heat required fr all five parts, and this will be the ttal amunt f heat required t cnvert 1.00 ml f H O(s) at!0.ec t H O(g) at 140.EC. q 1 =.0 J/ECCg 18.0 g [0 (!0.)]EC = 1.1 10 J q = 1.00 ml 6.0 10 J/ml = 6.0 10 J q = /ECCg 18.0 g (100. 0)EC = 7.5 10 J q 4 = 1.00 ml 40.7 10 J/ml = 4.07 10 4 J q 5 =.0 J/ECCg 18.0 g (140. 100.)EC = 1.5 10 J q ttal = q 1 + q + q + q 4 + q 5 = 5.69 10 4 J = 56.9 kj 14. When a mixture f ice and water exists, the temperature f the mixture remains at 0EC until all f the ice has melted. Because an ice-water mixture exists at the end f the prcess, the temperature remains at 0EC. All f the energy released by the element ges t cnvert ice int water. The energy required t d this is related t H fusin = 6.0 kj/ml (frm Exercise 1). Heat lss by element = heat gain by ice cubes at 0EC 1 ml H O 6.0 kj Heat gain = 109.5 g H O = 6.6 kj 18.0 g ml H O Specific heat f element = q 6,600 J = = 0.75 J/ECCg mass T 500.0 g (195 0) C

08 CHAPTER 6 THERMOCHEMISTRY Integrative Prblems 15. N (g) + O (g) NO (g) H = 67.7 kj n N = n O = PV RT PV RT.50 atm 0.50 L = =.86 0.0806 L atm 7 K K ml.50 atm 0.450 L = = 5.15 0.0806 L atm 7 K K ml 10 ml N 10 ml O The balanced equatin requires a : 1 O t N mle rati. The actual mle rati is 5.15 10 /.86 10 = 1.80; Because the actual mle rati is smaller than the required mle rati, O in the numeratr is limiting. 5.15 5.15 10 ml O 10 ml NO ml NO ml O 67.7 kj ml NO = 5.15 = 1.74 kj 10 ml NO 16. a. 4 CH NO (l) + O (g) 4 CO (g) + N (g) + 6 H O(g) H rxn =!188.5 kj = [4 ml(!9.5 kj/ml) + 6 ml(!4 kj/ml)]! Slving:, CH NO H =!44 kj/ml f [4 ml( H f, CHNO )] b. P ttal = 950. trr n N 1atm 760 trr = 1.5 atm; 0.168 atm 15.0 L = = 0.08 ml N 0.0806 L atm 7 K K ml P = χ = 1.5 atm 0.14 N P ttal N = 0.168 atm 0.08 ml N 8.0 g N 1 ml N =.1 g N 17. Heat lss by U = heat gain by heavy water; vlume f cube = (cube edge) Mass f heavy water = 1.00 10 ml 1.11g ml = 1110 g

CHAPTER 6 THERMOCHEMISTRY 09 Heat gain by heavy water = 4.11 J 1110 g (8.5 5.5)EC = 1.4 10 4 J Heat lss by U = 1.4 10 4 J = 0.117 J mass (00.0 8.5)EC, mass = 7.0 10 g U 7.0 10 g U 1cm 19.05 g = 7 cm ; cube edge = (7 cm ) 1/ =. cm Marathn Prblems 18. X CO (g) + H O(l) + O (g) + A(g) H = 189 kj/ml (unbalanced) T determine X, we must determine the mles f X reacted, the identity f A, and the mles f A prduced. Fr the reactin at cnstant P ( H = q): q = q = -4.184 J/ CCg(1.000 10 4 g)(9.5 5.00 C)(1 kj/1000 J) H O rxn q rxn = 189.1 kj (carrying extra significant figures) Because H = 189 kj/ml fr the decmpsitin reactin, and because nly 189.1 kj f heat was released fr this reactin, 189.1 kj (1 ml X/189 kj) = 0.100 ml X was reacted. Mlar mass f X =.7 g X 0.100 ml X = 7 g/ml Frm the prblem, 0.100 ml X prduced 0.00 ml CO, 0.50 ml H O, and 0.05 ml O. Therefre, 1.00 ml X cntains.00 ml CO,.50 ml H O, and 0.5 ml O. 44.0 g 18.0 g 1.00 ml X = 7 g =.00 ml CO +.50 ml H O ml ml.0 g + 0.5 ml O ml Mass f A in 1.00 ml X = 7 g 1 g 45.0 g 8.0 g = 4 g A + (mass f A) T determine A, we need the mles f A prduced. The ttal mles f gas prduced can be determined frm the gas law data prvided in the prblem. Because H O(l) is a prduct, we need t subtract P frm the ttal pressure. O H PV n ttal = ; Pttal = P gases + RT P H O ; P gases = 778 trr 1 trr = 747 trr

10 CHAPTER 6 THERMOCHEMISTRY V = height area; area = πr ; V = (59.8 cm)(π)(8.00 cm) 1 L 1000 cm T = 7.15 + 9.5 = 0.67 K = 1.0 L PV 747 1atm trr (1.0 L) 760 trr n ttal = = RT 0.0806 L atm (0.67 K) K ml = 0.475 ml = ml CO + ml O + ml A Ml A = 0.475 ml ttal 0.00 ml CO 0.05 ml O = 0.150 ml A Because 0.100 ml X reacted, 1.00 ml X wuld cntain 1.50 ml A, which frm a previus calculatin represents 4 g A. Mlar mass f A = 4 g A 1.50 ml A = 8 g/ml Because A is a gaseus element, the nly element that is a gas and has this mlar mass is N (g). Thus A = N (g). a. Nw we can determine the frmula f X. X CO (g) +.5 H O(l) + 0.5 O (g) + 1.5 N (g). Fr a balanced reactin, X = C H 5 N O 9, which, fr yur infrmatin, is nitrglycerine. 1atm b. w = P V = 778 trr (1.0 L 0) = 1. L atm 760 trr 8.145 J/K ml 1. L atm = 150 J = 1.5 kj, w = 1.5 kj 0.0806 L atm/k ml c. E = q + w, where q = H since at cnstant pressure. Fr 1 ml f X decmpsed: w = 1.5 kj/0.100 ml = 1.5 kj/ml E = H + w = 189 kj/ml + ( 1.5 kj/ml) = 1906 kj/ml H f assuming H Hrxn. 189 kj: fr C H 5 N O 9 can be estimated frm standard enthalpies f frmatin data and rxn = Fr the balanced reactin given in part a, where, CO f, H O H rxn = f, C H N O 189 kj = ( H +.5 H + 0.5 H + 1.5 H ) ( H ) f, O f, N f 5 9 189 kj = [( 9.5) kj +.5( 86) kj + 0 + 0] H, C H N O H =.5 kj/ml = kj/ml f 5 9, C H N O f 5 9

CHAPTER 6 THERMOCHEMISTRY 11 x + y/ 19. C x H y + Ο x CO + y/ H O [x(!9.5) + y/ (!4)]! H C xh y =!044.5,!(9.5)x! 11y! H Cx H y =!044.5 P MM d gas =, RT 0.751 g/l = where MM = average mlar mass f CO /H O mixture 1.00 atm MM, MM f CO /H O mixture = 9.1 g/ml 0.0806 L atm 47 K K ml Let a = ml CO and 1.00! a = ml H O (assuming 1.00 ttal mles f mixture) (44.01)a + (1.00! a) 18.0 = 9.1; slving: a = 0.46 ml CO ; ml H O = 0.574 ml Thus: 0.574 0.46 y =, x. 69 = y x, y = (.69)x Fr whle numbers, multiply by three, which gives y = 8, x =. Nte that y = 16, x = 6 is pssible, alng with ther cmbinatins. Because the hydrcarbn has a lwer density than Kr, the mlar mass f C x H y must be less than the mlar mass f Kr (8.80 g/ml). Only C H 8 wrks.!044.5 =!9.5()! 11(8)! H H, H C H C 8 8 =!104 kj/ml