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); the cmbustin f gasline is exthermic (as is typical f cmbustin reactins). Fr exthermic reactins, heat is released int the surrundings giving a negative q value. 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 because the mles f gas increase. When a gas expands, the system des wrk n the surrundings, and w is a negative value. 14. H = E + PV at cnstant P; frm the definitin f enthalpy, the difference between H and E, at cnstant P, is the quantity PV. Thus, when a system at cnstant P can d pressure-vlume 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. a. 1 ml Hg 90.7 kj ml Hg = 90.7 kj required 184

CHAPTER 6 THERMOCHEMISTRY 185 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; 181.4 kj released 17. Given: CH 4 (g) + O (g) CO (g) + H O(l) CH 4 (g) + O (g) CO (g) + H O(g) H = 891 kj H = 80 kj Using Hess s law: 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) The enthalpy f vaprizatin f water is 44 kj/ml. H = H 1 + H = 44 kj Nte: When an equatin is reversed, the sign n ΔH is reversed. When the cefficients in a balanced equatin are multiplied by an integer, then the value f ΔH is multiplied by the same integer. 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 Δ H f values are elements in their standard state. All substances are measured in relatinship t this zer pint. 0. a. CH 4 (g) + O (g) CO (g) + H O(l) H =? Utilizing Hess s law: Reactants Standard State Elements H = H a + H b = 75 + 0 = 75 kj Standard State Elements Prducts H = H c + H d = 94 57 = 966 kj Reactants Prducts H = 75 966 = 891 kj b. The standard enthalpy f frmatin fr an element in its standard state is given a value f zer. T assign standard enthalpy f frmatin values fr all ther substances, there needs t be a reference pint frm which all enthalpy changes are determined. This reference pint is the elements in their standard state which is defined as the zer pint.

186 CHAPTER 6 THERMOCHEMISTRY S when using standard enthalpy values, a reactin is brken up int tw steps. The first step is t calculate the enthalpy change necessary t cnvert the reactants t the elements in their standard state. The secnd step is t determine the enthalpy change that ccurs when the elements in their standard state g t frm the prducts. When these tw steps are added tgether, the reference pint (the elements in their standard state) cancels ut and we are left with the enthalpy change fr the reactin. c. This verall reactin is just the reverse f all the steps in the part a answer. S H = +966 75 = 891 kj. Prducts are first cnverted t the elements in their standard state which requires 966 kj f heat. Next, the elements in the standard states g t frm the riginal reactants [CH 4 (g) + O (g)] which has an enthalpy change f 75 kj. All f the signs are reversed because the entire prcess is reversed. 1. 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.. 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.. 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. 4. Advantages: H burns cleanly (less pllutin) and gives a lt f energy per gram f fuel. Water as a surce f hydrgen is abundant and cheap. Disadvantages: Expensive and gas strage and safety issues Exercises Ptential and Kinetic Energy 5. KE = 1 mv ; cnvert mass and velcity t SI units. 1 J = Mass = 5.5 z 1lb 16z 1kg = 0.149 kg.05lb 1 kg m s Velcity = 1.0 10 h mi 1h 60min 1min 60s 1760yd mi 1m 1.094yd 45m s KE = 1 mv = 1 45 m 0.149 kg s = 150 J

CHAPTER 6 THERMOCHEMISTRY 187 1 6. KE = 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. 7. a. Ptential energy is energy due t psitin. Initially, ball A has a higher ptential energy than ball B because the psitin f ball A is higher than the psitin f ball B. In the final psitin, ball B has the higher psitin s ball B has the higher ptential energy. b. As ball A rlled dwn the hill, sme f the ptential energy lst by A has been cnverted t randm mtin f the cmpnents f the hill (frictinal heating). The remainder f the lst ptential energy was added t B t initially increase its kinetic energy and then t increase its ptential energy. 9.81m 8. Ball A: PE = mgz =.00 kg s 196kg 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. At pint II: PE = mgz = 4.00 kg 9.81m s KE = E ttal PE = 196 J 118 J = 78 J.00 m = 118 J Heat and Wrk 9. ΔE = q + w = 45 kj + (9 kj) = 16 kj 0. ΔE = q + w = 15 + 104 = 1 kj 1. a. ΔE = q + w = 47 kj + 88 kj = 41 kj b. ΔE = 8 47 = 5 kj c. ΔE = 47 + 0 = 47 kj d. When the surrundings d wrk n the system, w > 0. This is the case fr a.. Step 1: ΔE 1 = q + w = 7 J + 5 J = 107 J; step : ΔE = 5 J 7 J = 7 J ΔE verall = ΔE 1 + ΔE = 107 J 7 J = 70. J. ΔE = 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

188 CHAPTER 6 THERMOCHEMISTRY 4. a. ΔE = q + w = J + 100. J = 77 J b. w = PΔV = 1.90 atm(.80 L 8.0 L) = 10.5 L atm ΔE = q + w = 50. J + 1060 = 1410 J c. w = PΔV = 1.00 atm(9.1 L11. L) = 17.9 L atm 101. J L atm 101. J L atm = 1060 J = 1810 J ΔE = q + w = 107 J 1810 J = 770 J 5. w = PΔV; 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.00atm w = PΔV =.00 atm(75.0 L 10.0 L) = 10. L atm 6. w = 10. J = PΔV, 10 J = P(5 L 10. L), P = 14 atm 7. In this prblem, q = w = 950. J. 950. J 1 L atm = 9.8 L atm f wrk dne by the gases 101. J 101. J L atm 1kJ 1000J = 1. kj = wrk w = PΔV, 9.8 L atm = 650. 760 8. ΔE = 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 = PΔV, 1.50 L atm = 0.500 atm ΔV, ΔV =.06 L ΔV = V f V i,.06 L = 58.0 L V i, V i = 54.9 L = initial vlume 9. q = mlar heat capacity ml ΔT = w = PΔV = 1.00 atm (998 L 876 L) = 1 L atm ΔE = q + w = 0.9 kj + (1.4 kj) = 18.5 kj = 1.50 L atm 0.8 J 9.1 ml (8.0 0.0) C = 0,900 J C ml = 0.9 kj 101. J L atm = 1,400 J = 1.4 kj

CHAPTER 6 THERMOCHEMISTRY 189 40. H O(g) H O(l); ΔE = q + w; q = 40.66 kj; w = PΔV Vlume f 1 ml H O(l) = 1.000 ml H O(l) 18.0g ml 1cm = 18.1 cm = 18.1 ml 0.996g w = PΔV = 1.00 atm (0.0181 L 0.6 L) = 0.6 L atm ΔE = q + w = 40.66 kj +.10 kj = 7.56 kj 101. J L atm =.10 10 J =.10 kj Prperties f Enthalpy 41. 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. 4. 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. 4. 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. 44. 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 vapr 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. 45. 4 Fe(s) + O (g) Fe O (s) ΔH = 165 kj; nte that 165 kj f heat is released when 4 ml Fe reacts with ml O t prduce ml Fe O. 165kJ a. 4.00 ml Fe = 1650 kj; 1650 kj f heat released 4 mlfe 165kJ b. 1.00 ml Fe O = 86 kj; 86 kj f heat released mlfe O

190 CHAPTER 6 THERMOCHEMISTRY 1mlFe 165kJ c. 1.00 g Fe = 7.9 kj; 7.9 kj f heat released 55.85g 4 mlfe d. 10.0 g Fe 1 ml Fe 165kJ = 7.9 kj 55.85g Fe 4 ml Fe.00 g O 1 mlo 165kJ = 4.4 kj.00g O mlo Because.00 g O releases the smaller quantity f heat, O is the limiting reactant and 4.4 kj f heat can be released frm this mixture. 46. a. 1.00 ml H O 57kJ mlh O = 86 kj; 86 kj f heat released b. 4.0 g H c. 186 g O 1mlH 57kJ = 57 kj; 57 kj f heat released.016g H mlh 1 ml O 57kJ = 0 kj; 0 kj f heat released.00g O ml O PV d. n H = = RT 1.0 atm.0 10 L 0.0806L atm 98K K ml 8 = 8. 10 6 ml H 8. 10 6 ml H 57kJ ml H =. 10 9 kj;. 10 9 kj f heat released 47. 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 1mlC H 8 1 10 J 44.09g CH mlc H 8 8 = 4.4 10 g C H 8 48. a. 1.00 g CH 4 1 ml CH 4 891kJ = 55.5 kj 16.04g CH mlch 4 4 b. n = PV = RT 740. trr 1atm 760trr 0.0806L atm K ml 1.00 10 98K L = 9.8 ml CH 4 891kJ 9.8 ml CH 4 =.55 10 4 kj ml CH 4 49. 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.

CHAPTER 6 THERMOCHEMISTRY 191 50. H = E + PV; 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 51. Specific heat capacity is defined as the amunt f heat necessary t raise the temperature f ne gram f substance by ne 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 ΔT = 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: ΔT = energy s m 10.7 kj 0.14J 1000J kj 550. g = 140 C 5. a. s = specific heat capacity = 0.4J 0.4J since ΔT(K) = ΔT( C) K g Energy = s m ΔT = 0.4J 150.0 g (98 K 7 K) = 9.0 10 J b. Mlar heat capacity = 0.4J 107.9 g Ag mlag 6J C ml c. 150 J = 0.4J m (15. C 1.0 C), m = 150 0.4. = 1.6 10 g Ag 5. s = specific heat capacity = q m T 1J = 0.890 J/ C g 5.00g (55.1 5.) C Frm Table 6.1, the substance is slid aluminum.

19 CHAPTER 6 THERMOCHEMISTRY 54. s = 585J 15.6 g (5.5 0.0) C = 0.19 J/ C g Mlar heat capacity = 0.19J 00.6 g mlhg 7.9 J C ml 55. 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 = 4.18 J/ C g = 4.18 J/K g (ΔT in C = ΔT in K). Heat lss by ht water = s m ΔT = 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. 56. 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 = 90.0g 15.0 C 18.0 C = 75.0 g ht water needed 57. Heat lss by Al + heat lss by Fe = heat gain by water; keeping all quantities psitive t avid sign errr: 0.89J 5.00 g Al (100.0 C T f ) + 0.45J 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

CHAPTER 6 THERMOCHEMISTRY 19 58. Heat released t water = 5.0 g H 10.J g H + 10. g methane 50.J = 1.10 10 J g methane 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 59. Heat gain by water = heat lss by metal = s m ΔT, 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 = 100J 150.0 g 56.7 = 0.5 J/ C g C 60. Heat gain by water = heat lss by Cu; keeping all quantities psitive helps t avid sign errrs: mass (4.9 C. C) = 0.0J 110. g Cu (8.4 C 4.9 C) 11(mass) = 100, mass = 10 g H O 61. 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 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 ΔH (heat per ml AgCl frmed) is negative with a value f: ΔH = 0J 5.00 10 ml 1kJ 1000J = 66 kj/ml Nte: Sign errrs are cmmn with calrimetry prblems. Hwever, the crrect sign fr ΔH can be determined easily frm the ΔT data; i.e., if ΔT f the slutin increases, then the reactin is exthermic because heat was released, and if ΔT 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 ΔH. This will help eliminate sign errrs.

194 CHAPTER 6 THERMOCHEMISTRY 6. 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 = 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 ΔH 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 ΔH. This will help eliminate sign errrs. ΔH = 5.6 kj 0.10ml = 56 kj/ml 6. 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 ΔH can easily be btained frm the ΔT data. When wrking calrimetry prblems, keep all quantities psitive (ignre signs). When finished, deduce the crrect sign fr ΔH. Fr this prblem, T decreases as KBr disslves, s ΔH 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 ΔH in units f J/g = 1800J 10.5 g KBr = 170 J/g ΔH in units f kj/ml = 170J g KBr 119.0 g KBr ml KBr 1kJ 1000J = 0. kj/ml 64. NH 4 NO (s) NH 4 + (aq) + NO (aq) ΔH =?; 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 ΔT) and then deduce the crrect sign fr ΔH 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 (ΔH is psitive). Heat lst by slutin = 76.6 g (5.00.4) C = 5 J = heat gained as NH 4 NO disslves

CHAPTER 6 THERMOCHEMISTRY 195 ΔH = 5J 1.60g NH NO 4 80.05g NH4NO 1kJ = 6.6 kj/ml NH 4 NO disslving ml NH NO 1000J 4 65. Because ΔH is exthermic, the temperature f the slutin will increase as CaCl (s) disslves. Keeping all quantities psitive: 66. 0.1000 L 1ml CaCl 81.5 kj heat lss as CaCl disslves = 11.0 g CaCl = 8.08 kj 110.98g CaCl ml CaCl heat gained by slutin = 8.08 10 J = T f 5.0 C = 8.08 4.18 0.500ml HCl L (15 + 11.0) g (T f 5.0 C) 10 = 14. C, T f = 14. C + 5.0 C = 9. C 16 118kJ heat released =.95 kj f heat released if HCl limiting mlhcl 0.000 L 0.100ml Ba(OH) L 118kJ heat released ml Ba(OH) =.54 kj heat released if Ba(OH) limiting Because the HCl reagent prduces the smaller amunt f heat released, HCl is limiting and.95 kj f heat are released by this reactin. Heat gained by slutin =.95 10 J = ΔT = 1.76 C = T f T i = T f 5.0 C, T f = 6.8 C 400.0 g ΔT 67. a. Heat gain by calrimeter = heat lss by CH 4 = 6.79 g CH 4 1mlCH 4 80kJ 16.04g 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 A bmb calrimeter is at cnstant vlume, s the heat released/gained = q V = E: 5kJ ΔE cmb = 1.6 g C H 6.04g mlc H = 1.10 10 kj/ml 68. 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. Heat lss = 0.1584 g 6.4kJ g = 4.185 kj

196 CHAPTER 6 THERMOCHEMISTRY Heat gain = 4.185 kj = C cal ΔT, C cal = 4.185kJ.54 C = 1.65 kj/ C Nw we can calculate the heat f cmbustin f vanillin. Heat lss = heat gain. 1.65kJ 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 = 5.6kJ 0.10g = 5. kj/g; E cmb = 5. kj g 15.14g ml = 80 kj/ml Hess's Law 69. Infrmatin given: C(s) + O (g) CO (g) CO(g) + 1/ O (g) CO (g) ΔH = 9.7 kj ΔH = 8. kj 70. Given: Using Hess s law: C(s) + O (g) CO (g) ΔH 1 = (9.7 kj) CO (g) CO(g) + O (g) ΔH = (8. kj) C(s) + O (g) CO(g) ΔH = ΔH 1 + ΔH = 0.8 kj Nte: When an equatin is reversed, the sign n ΔH is reversed. When the cefficients in a balanced equatin are multiplied by an integer, then the value f ΔH is multiplied by the same integer. C 4 H 4 (g) + 5 O (g) 4 CO (g) + H O(l) C 4 H 8 (g) + 6 O (g) 4 CO (g) + 4 H O(l) H (g) + 1/ O (g) H O(l) ΔH cmb = 41 kj ΔH cmb = 755 kj ΔH 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) ΔH 1 = 41 kj 4 CO (g) + 4 H O(l) C 4 H 8 (g) + 6 O (g) ΔH = (755 kj) H (g) + O (g) H O(l) ΔH = (86 kj) C 4 H 4 (g) + H (g) C 4 H 8 (g) ΔH = ΔH 1 + ΔH + ΔH = 158 kj

CHAPTER 6 THERMOCHEMISTRY 197 71. N (g) + 6 H (g) 4 NH (g) ΔH = (9 kj) 6 H O(g) 6 H (g) + O (g) ΔH = (484 kj) N (g) + 6 H O(g) O (g) + 4 NH (g) ΔH = 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. 7. ClF + 1/ O 1/ Cl O + 1/ F O ΔH = 1/(167.4 kj) 1/ Cl O + / F O ClF + O ΔH = 1/(41.4 kj) F + 1/ O F O ΔH = 1/(4.4 kj) ClF(g) + F (g) ClF ΔH = 108.7 kj 7. NO + O NO + O ΔH = 199 kj / O O ΔH = 1/(47 kj) O 1/ O ΔH = 1/(495 kj) NO(g) + O(g) NO (g) ΔH = kj 74. We want ΔH fr N H 4 (l) + O (g) N (g) + H O(l). It will be easier t calculate ΔH fr the cmbustin f fur mles f N H 4 because we will avid fractins. 9 H + 9/ O 9 H O ΔH = 9(86 kj) N H 4 + H O N O + 9 H ΔH = (17 kj) NH + N O 4 N + H O ΔH = 1010. kj N H 4 + H O NH + 1/ O ΔH = (14 kj) 4 N H 4 (l) + 4 O (g) 4 N (g) + 8 H O(l) ΔH = 490. kj 490. kj Fr N H 4 (l) + O (g) N (g) + H O(l) ΔH = = 6 kj 4 Nte: By the significant figure rules, we culd reprt this answer t fur significant figures. Hwever, because the ΔH values given in the prblem are nly knwn t ±1 kj, ur final answer will at best be ±1 kj. 75. CaC Ca + C ΔH = (6.8 kj) CaO + H O Ca(OH) ΔH = 65.1 kj CO + H O C H + 5/ O ΔH = (100. kj) Ca + 1/ O CaO ΔH = 65.5 kj C + O CO ΔH = (9.5 kj) CaC (s) + H O(l) Ca(OH) (aq) + C H (g) ΔH = 71 kj

198 CHAPTER 6 THERMOCHEMISTRY 76. P 4 O 10 P 4 + 5 O ΔH = (967. kj) 10 PCl + 5 O 10 Cl PO ΔH = 10(85.7 kj) 6 PCl 5 6 PCl + 6 Cl ΔH = 6(84. kj) P 4 + 6 Cl 4 PCl ΔH = 15.6 P 4 O 10 (s) + 6 PCl 5 (g) 10 Cl PO(g) ΔH = 610.1 kj Standard Enthalpies f Frmatin 77. 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) 78. 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: ΔH cmb values assume 1 mle f cmpund cmbusted. f. NH 4 Br(s) NH 4 + (aq) + Br (aq) 79. In general, ΔH = n p ΔH f n, prducts r ΔH f, and all elements in their standard, reactants state have Δ = 0 by definitin. H f a. The balanced equatin is NH (g) + O (g) + CH 4 (g) HCN(g) + 6 H O(g). ΔH = ( ml HCN ΔH f, HCN + 6 ml H O(g) H f, HO Δ ) ( ml NH Δ + ml CH 4 H ) H f, NH f, CH 4 ΔH = [(15.1) + 6(4)] [(46) + (75)] = 940. kj

CHAPTER 6 THERMOCHEMISTRY 199 b. Ca (PO 4 ) (s) + H SO 4 (l) CaSO 4 (s) + H PO 4 (l) ΔH = 14kJ 167kJ mlcaso 4 (s) mlhpo4(l) ml ml 416kJ mlca (PO4) (s) mlhso ml 1 4 814kJ (l) ml ΔH = 68 kj (6568 kj) = 65 kj c. NH (g) + HCl(g) NH 4 Cl(s) ΔH = (1 ml NH 4 Cl Δ ) (1 ml NH H f, NH4Cl ΔH f, + 1 ml HCl H f, HCl NH Δ ) ΔH = 14kJ 1ml ml 46kJ 9kJ 1ml 1ml ml ml ΔH = 14 kj + 18 kj = 176 kj 80. a. The balanced equatin is C H 5 OH(l) + O (g) CO (g) + H O(g). ΔH = 9.5 kj 4kJ ml ml ml ml 78kJ 1ml ml ΔH = 151 kj (78 kj) = 15 kj b. SiCl 4 (l) + H O(l) SiO (s) + 4 HCl(aq) Because HCl(aq) is H + (aq) + Cl (aq), Δ H f = 0 167 = 167 kj/ml. ΔH = 167kJ 911kJ 4 ml 1ml ml ml 687kJ 86kJ 1ml ml ml ml ΔH = 1579 kj (159 kj) = 0. kj c. MgO(s) + H O(l) Mg(OH) (s) ΔH = 95kJ 1 ml ml 60kJ 1ml ml 86kJ 1ml ml ΔH = 95 kj (888 kj) = 7 kj 81. a. 4 NH (g) + 5 O (g) 4 NO(g) + 6 H O(g); ΔH = n p ΔH f n, prducts r ΔH f, reactants ΔH = 90. kj 4kJ 46kJ 4 ml 6 ml 4 ml = 908 kj ml ml ml

00 CHAPTER 6 THERMOCHEMISTRY NO(g) + O (g) NO (g) ΔH = 4kJ 90. kj ml ml = 11 kj ml ml NO (g) + H O(l) HNO (aq) + NO(g) ΔH = Nte: All 07kJ 90. kj ml 1ml ml ml ΔH f values are assumed ±1 kj. 4kJ 86kJ ml 1ml 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. 416kJ 8. 4 Na(s) + O (g) Na O(s) ΔH = ml ml = 8 kj Na(s) + H O(l) NaOH(aq) + H (g) ΔH = 470. kj 86kJ ml ml = 68 kj ml ml Na(s) + CO (g) Na O(s) + CO(g) ΔH = 416kJ 110.5 kj 9.5 kj 1 ml 1ml 1ml = 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. 8. Al(s) + NH 4 ClO 4 (s) Al O (s) + AlCl (s) + NO(g) + 6 H O(g) ΔH = 4kJ 90. kj 704kJ 1676kJ 6 ml ml 1ml 1ml ml ml ml ml 95kJ ml = 677 kj ml

CHAPTER 6 THERMOCHEMISTRY 01 84. 5 N O 4 (l) + 4 N H CH (l) 1 H O(g) + 9 N (g) + 4 CO (g) ΔH = 4kJ 9.5 kj 1 ml 4 ml ml ml 0. kj 54kJ 5 ml 4 ml = 4594 kj ml ml 85. ClF (g) + NH (g) N (g) + 6 HF(g) + Cl (g) ΔH = 1196 kj H f, HF, ClF f, NH ΔH = (6 Δ ) ( H H ) f 71kJ 46kJ 1196 kj = 6 ml ΔH f, ClF ml ml ml H f, ClF 1196 kj = 166 kj Δ + 9 kj, H f, ClF Δ = 86. C H 4 (g) + O (g) CO (g) + H O(l) ΔH = 1411.1 kj ΔH = 1411.1 kj = (9.5) kj + (85.8) kj ΔH f, ( 166 9 1196) kj = ml C H 4 169kJ ml 1411.1 kj = 158.6 kj ΔH, f, C H 4 ΔH f, C H 4 = 5.5 kj/ml Energy Cnsumptin and Surces 87. C(s) + H O(g) H (g) + CO(g) ΔH = 110.5 kj (4 kj) = 1 kj 88. CO(g) + H (g) CH OH(l) ΔH = 9 kj (110.5 kj) = 19 kj 89. C H 5 OH(l) + O (g) CO (g) + H O(l) ΔH = [(9.5 kj) + (86 kj)] (78 kj) = 167 kj/ml ethanl 167kJ ml 1ml 46.07g = 9.67 kj/g 90. CH OH(l) + / O (g) CO (g) + H O(l) ΔH = [9.5 kj + (86 kj)] (9 kj) = 77 kj/ml CH OH 77kJ ml 1ml.04g =.7 kj/g versus 9.67 kj/g fr ethanl (frm Exercise 89) Ethanl has a slightly higher fuel value per gram than methanl.

0 CHAPTER 6 THERMOCHEMISTRY 91. C H 8 (g) + 5 O (g) CO (g) + 4 H O(l) ΔH = [(9.5 kj) + 4(86 kj)] (104 kj) = 1 kj/ml C H 8 1kJ ml 1ml 44.09g = 50.7kJ g 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. 9. 1 mle f C H (g) and 1 mle f C 4 H 10 (g) have equivalent vlumes at the same T and P. Enthalpyf cmbustin per vlumef Enthalpyf cmbustin per vlumef CH C H 4 10 = enthalpyf cmbustin per mlf enthalpyf cmbustin per mlf C C 4 H H 10 Enthalpyf cmbustin per vlumef Enthalpyf cmbustin per vlumef CH C H 4 10 = 49.9 kj 6.04g CH g CH mlch 49.5 kj 58.1g C4H g C H mlc 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. 9. The mlar vlume f a gas at STP is.4 L (frm Chapter 5). 4.19 10 6 1mlCH kj 891kJ 94. Mass f H O = 1.00 gal 4.4L CH 4 = 1.05 10 5 L CH 4 mlch.785l gal 4 1000mL 1.00g = 790 g H O L ml Energy required (theretical) = s m ΔT = 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 J 100. J 80.0 J = 1.98 10 5 J Mass f C H = 1.98 10 5 J 1mlC H 100. 10 J 6.04g CH mlc H =.97 g C H Additinal Exercises 95..0 h 5500kJ 1mlHO 18.0g HO h 40.6 kj ml = 4900 g = 4.9 kg H O

CHAPTER 6 THERMOCHEMISTRY 0 96. Frm the prblem, walking 4.0 miles cnsumes 400 kcal f energy. 1 lb fat 454g lb 7.7 kcal g 4 mi 400kcal 1h 4 mi = 8.7 h = 9 h 97. a. SO (g) + O (g) SO (g); w = PΔV; 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 (ΔV 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 (ΔV 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), ΔV = 0 and w = 0. 98. a. N (g) + H (g) NH (g); frm the balanced equatin, 1 mlecule f N will react with mlecules f H t prduce mlecules f NH. S the picture after the reactin shuld nly have mlecules f NH present. Anther imprtant part f yur drawing will be the relative vlume f the prduct cntainer. The vlume f a gas is directly prprtinal t the number f gas mlecules present (at cnstant T and P). In this prblem, 4 ttal mlecules f gas were present initially (1 N + H ). After reactin, nly mlecules are present ( NH ). Because the number f gas mlecules decreases by a factr f (frm 4 ttal t ttal), the vlume f the prduct gas must decrease by a factr f as cmpared t the initial vlume f the reactant gases. Summarizing, the picture f the prduct cntainer shuld have mlecules f NH and shuld be at a vlume which is ne-half the riginal reactant cntainer vlume. b. w = PV; here the vlume decreased, s V is negative. When V is negative, w is psitive. As the reactants were cnverted t prducts, a cmpressin ccurred which is assciated with wrk flwing int the system (w is psitive). 99. a. C 1 H O 11 (s) + 1 O (g) 1 CO (g) + 11 H O(l) b. A bmb calrimeter is at cnstant vlume, s heat released = q V = ΔE: ΔE = 4.00kJ 1.46g 4.0g ml = 560 kj/ml C 1 H O 11 c. PV = nrt; at cnstant P and T, PΔV = RTΔn, where Δn = mles f gaseus prducts mles f gaseus reactants. At cnstant P and T: ΔH = ΔE + PΔV = ΔE + RTΔn Fr this reactin, Δn = 1 1 = 0, s ΔH = ΔE = 560 kj/ml.

04 CHAPTER 6 THERMOCHEMISTRY 100. w = PΔV; Δn = 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 (Δn psitive), the reactin will expand in vlume (ΔV psitive), and the system will d wrk n the surrundings. Fr example, in reactin c, Δn = 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 (Δn negative), the reactin will cntract in vlume (ΔV negative), and the surrundings will d cmpressin wrk n the system, e.g., reactin a, where Δn = 0 1 = 1. When there is n change in the mles f gas frm reactants t prducts, ΔV = 0 and w = 0, e.g., reactin b, where Δn = = 0. When ΔV > 0 (Δn > 0), then w < 0, and the system des wrk n the surrundings (c and e). When ΔV < 0 (Δn < 0), then w > 0, and the surrundings d wrk n the system (a and d). When ΔV = 0 (Δn = 0), then w = 0 (b). 101. ΔE verall = ΔE step 1 + ΔE step ; this is a cyclic prcess, which means that the verall initial state and final state are the same. Because ΔE is a state functin, ΔE verall = 0 and ΔE step 1 = ΔE step. ΔE step 1 = q + w = 45 J + (10. J) = 5 J ΔE step = ΔE step 1 = 5 J = q + w, 5 J = 60 J + w, w = 5 J 10. K(s) + H O(l) KOH(aq) + H (g) ΔH = (481 kj) (86 kj) = 90. kj 5.00 g K 1mlK 90. kj = 4.9 kj 9.10g K mlk 4.9 kj f heat is released n reactin f 5.00 g K. 4,900 J = (1.00 10 4,900 g) ΔT, ΔT = g C 4.18 1.0010 = 5.96 C Final temperature = 4.0 + 5.96 = 0.0 C 10. HNO (aq) + KOH(aq) H O(l) + KNO (aq) ΔH = 56 kj 0.400ml HNO 0.000 L L 0.500ml KOH 0.1500 L L 56 kj heat released ml HNO 56 kj heat released ml KOH = 4.5 kj heat released if HNO limiting = 4. kj heat released if KOH limiting Because the KOH reagent prduces the smaller quantity f heat released, KOH is limiting and 4. kj f heat released.

CHAPTER 6 THERMOCHEMISTRY 05 104. Na SO 4 (aq) + Ba(NO ) (aq) BaSO 4 (s) + NaNO (aq) ΔH =? 1.00 L.00 ml Na L SO 4 1 ml BaSO 4 =.00 ml BaSO 4 if Na SO 4 limiting ml Na SO 4 0.750ml Ba(NO).00 L L 1mlBaSO 4 = 1.50 ml BaSO 4 if Ba(NO ) limiting ml Ba(NO ) The Ba(NO ) reagent prduces the smaller quantity f prduct, s Ba(NO ) is limiting and 1.50 ml BaSO 4 can frm. Heat gain by slutin = heat lss by reactin 1000mL Mass f slutin =.00 L L.00g = 6.00 10 g ml Heat gain by slutin = 6.7 J 6.00 10 g (4.0 0.0)C = 4.59 10 5 J Because the slutin gained heat, the reactin is exthermic; q = 4.59 10 5 J fr the reactin. 4.59 10 J H = 1.50 ml BaSO 5 4 =.06 10 5 J/ml = 06 kj/ml 105. 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. 106. The specific heat f water is 4.18 J/ C g, which is equal t 4.18 kj/ C kg. We have 1.00 kg f H O, s: 1.00 kg 4.18 kj = 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 107. Heat released = 1.056 g 6.4 kj/g = 7.90 kj = heat gain by water and calrimeter Heat gain = 7.90 kj = 4.18kJ C kg 0.987kg ΔT 6.66 kj C ΔT 7.90 = (4.1 + 6.66)ΔT = (10.79)ΔT, ΔT =.586 C.586 C = T f. C, T f = 5.91 C

06 CHAPTER 6 THERMOCHEMISTRY 108. Fr Exercise 8, a mixture f ml Al and ml NH 4 ClO 4 yields 677 kj f energy. The mass f the stichimetric reactant mixture is: 6.98g 117.49g ml ml = 4.41 g ml ml Fr 1.000 kg f fuel: 1.000 10 g 677kJ 4.41g = 6177 kj In Exercise 84, we get 4594 kj f energy frm 5 ml f N O 4 and 4 ml f N H CH. The 9.0g 46.08g mass is 5 ml 4 ml = 644.4 kj. ml ml Fr 1.000 kg f fuel: 1.000 10 g 4594kJ 644.4g = 719 kj Thus we get mre energy per kilgram frm the N O 4 /N H CH mixture. 109. 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 110. T avid fractins, let's first calculate ΔH fr the reactin: 6 FeO(s) + 6 CO(g) 6 Fe(s) + 6 CO (g) 6 FeO + CO Fe O 4 + CO ΔH = (18 kj) Fe O 4 + CO Fe O + CO ΔH = (9 kj) Fe O + 9 CO 6 Fe + 9 CO ΔH = ( kj) 6 FeO(s) + 6 CO(g) 6 Fe(s) + 6 CO (g) ΔH = 66 kj S fr FeO(s) + CO(g) Fe(s) + CO (g), ΔH = 111. a. ΔH = ml(7 kj/ml) 1 ml(49 kj/ml) = 6 kj 66kJ = 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. Nte that mles f C H has the same mass as 1 mle f C 6 H 6.

CHAPTER 6 THERMOCHEMISTRY 07 11. I(g) + Cl(g) ICl(g) ΔH = (11. kj) 1/ Cl (g) Cl(g) ΔH = 1/(4. kj) 1/ I (g) I(g) ΔH = 1/(151.0 kj) 1/ I (s) 1/ I (g) ΔH = 1/(6.8 kj) 1/ I (s) + 1/ Cl (g) ICl(g) ΔH = 16.8 kj/ml = ΔH f, ICl 11. Heat gained by water = heat lst by nickel = s m ΔT, where s = specific heat capacity. Heat gain = 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 = 0.444J mass (99.8 5.0) C, mass = 940 0.444 74.8 = 8 g 114. Heat gain by calrimeter = 1.56kJ. 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. ΔE cmb = 5.0 kj 0.1964g = 5 kj/g; ΔE cmb = 5kJ g 108.09g ml = 700 kj/ml 115. a. C H 4 (g) + O (g) CH CHO(g) + O (g) ΔH = 166 kj [14 kj + 5 kj] = 61 kj b. O (g) + NO(g) NO (g) + O (g) ΔH = 4 kj [90. kj + 14 kj] = 199 kj c. SO (g) + H O(l) H SO 4 (aq) ΔH = 909 kj [96 kj + (86 kj)] = 7 kj d. NO(g) + O (g) NO (g) ΔH = (4) kj (90.) kj = 11 kj ChemWrk Prblems The answers t the prblems 116-1 (r a variatin t these prblems) are fund in OWL. These prblems are als assignable in OWL. Challenge Prblems 14. 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).

08 CHAPTER 6 THERMOCHEMISTRY Pathway 1: w = PΔV =.00 atm(0.0 L 10.0 L) = 40.0 L atm Pathway : w = PΔV = 1.00 atm(0.0 L 10.0 L) = 0.0 L atm 101. J L atm 101. J L atm = 4.05 10 J =.0 10 J Nte: The sign is negative 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. 15. A(l) A(g) ΔH vap = 0.7 kj; at cnstant pressure, ΔH = q p = 0.7 kj Because PV = nrt, at cnstant pressure and temperature: w = PΔV = RTΔn, where: Δn = mles f gaseus prducts mles f gaseus reactants = 1 0 = 1 w = RTΔn = 8.145 J/K ml(80. + 7 K)(1 ml) = 940 J =.94 kj ΔE = q + w = 0.7 kj + (.94 kj) = 7.8 kj 16. Energy needed = 0. 10 g C h 1 H O 11 1mlC1HO11 4. g C H O Energy frm sun = 1.0 kw/m = 1000 W/m 1000J 1.0 kj = s m s m 10,000 m 1.0 kj s m 60s min 60min h 1 =.6 10 7 kj/h 11 5640kJ ml =. 10 5 kj/h Percent efficiency = energy used per hur ttalenergy per hur 100 =..6 5 10 7 10 kj 100 = 0.9% kj 17. Energy used in 8.0 hurs = 40. kwh = Energy frm the sun in 8.0 hurs = 10. kj s m 40. kj h s 600s = 1.4 10 5 kj h 60s 60min 8.0 h =.9 10 4 kj/m min h Only 19% f the sunlight is cnverted int electricity: 0.19 (.9 10 4 kj/m ) area = 1.4 10 5 kj, area = 5 m 18. a. HNO (aq) + Na CO (s) NaNO (aq) + H O(l) + CO (g) ΔH = [(467 kj) + (86 kj) + (9.5 kj)] [(07 kj) + (111 kj)] = 69 kj

CHAPTER 6 THERMOCHEMISTRY 09.0 10 4 gallns 1.1 10 8 g slutin 7.7 10 7 g HNO 7.7 10 7 g HNO 4 qt gal 946mL 1.4g = 1.1 10 8 g f cncentrated nitric qt ml acid slutin 70.0 g HNO = 7.7 10 7 g HNO 100.0 g slutin 1 mlhno 6.0g HNO 4. 10 7 kj f heat was released. 1 ml Na CO ml HNO 1 ml HNO 69kJ = 4. 10 7 kj 6.0g HNO mlhno 105.99g Na CO ml Na CO = 6.5 10 7 g Na CO b. They feared the heat generated by the neutralizatin reactin wuld vaprize the unreacted nitric acid, causing widespread airbrne cntaminatin. 19. 400 kcal 4.18kJ kcal = 1.7 10 kj 10 kj 1kg 9.81m.54cm 1m PE = mgz = 180lb 8 in = 160 J 00 J.05lb s in 100cm 00 J f energy is needed t climb ne step. The ttal number f steps t climb are: 10 6 J 1step 00J = 1 10 4 steps H f 10. H (g) + 1/ O (g) H O(l) ΔH = Δ, H O(l) = 85.8 kj; we want the reverse reactin: H O(l) H (g) + 1/ O (g) ΔH = 85.8 kj w = PV; because PV = nrt, at cnstant T and P, PV = RTn, where n = mles f gaseus prducts mles f gaseus reactants. Here, Δn = (1 ml H + 0.5 ml O ) (0) = 1.50 ml. ΔE = ΔH PΔV = ΔH RTΔn ΔE = 85.8 kj 8.145 J/K ml 98K 1 kj 1000J 1.50ml ΔE = 85.8 kj.7 kj = 8.1 kj 11. There are five parts t this prblem. We need t calculate: (1) q required t heat H O(s) frm 0. C t 0C; use the specific heat capacity f H O(s) () q required t cnvert 1 ml H O(s) at 0C int 1 ml H O(l) at 0C; use H fusin

10 CHAPTER 6 THERMOCHEMISTRY () q required t heat H O(l) frm 0C t 100.C; use the specific heat capacity f H O(l) (4) q required t cnvert 1 ml H O(l) at 100.C int 1 ml H O(g) at 100.C; use H vaprizatin (5) q required t heat H O(g) frm 100.C t 140.C; 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.C t H O(g) at 140.C. q 1 =.0 J/C g 18.0 g [0 (0.)]C = 1.1 10 J q = 1.00 ml 6.0 10 J/ml = 6.0 10 J q = 4.18 J/C g 18.0 g (100. 0)C = 7.5 10 J q 4 = 1.00 ml 40.7 10 J/ml = 4.07 10 4 J q 5 =.0 J/C g 18.0 g (140. 100.)C = 1.5 10 J q ttal = q 1 + q + q + q 4 + q 5 = 5.69 10 4 J = 56.9 kj 1. When a mixture f ice and water exists, the temperature f the mixture remains at 0C until all f the ice has melted. Because an ice-water mixture exists at the end f the prcess, the temperature remains at 0C. 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 11). Heat lss by element = heat gain by ice cubes at 0C Heat gain = 109.5 g H O 1mlHO 6.0kJ = 6.6 kj 18.0g mlh O Specific heat f element = q mass ΔT 6,600J = 0.75 J/C g 500.0 g (195 0) C Integrative Prblems 1. N (g) + O (g) NO (g) H = 67.7 kj n N = n O = PV RT PV RT.50atm 0.50L =.86 0.0806 L atm 7K K ml.50atm 0.450L = 5.15 0.0806 L atm 7K K ml 10 ml N 10 ml O

CHAPTER 6 THERMOCHEMISTRY 11.86 10 ml N ml NO 1 ml N = 5.7 10 ml NO prduced if N is limiting. 5.15 10 ml O ml NO mlo = 5.15 10 ml NO prduced if O is limiting. O is limiting because it prduces the smaller quantity f prduct. The heat required is: 5.15 10 ml NO 67.7 kj ml NO = 1.74 kj 14. a. 4 CH NO (l) + O (g) 4 CO (g) + N (g) + 6 H O(g) Δ H rxn = 188.5 kj = [4 ml(9.5 kj/ml) + 6 ml(4 kj/ml)] Slving:, CH NO Δ H = 44 kj/ml f [4 ml( H, CH NO f )] b. P ttal = 950. trr 1atm 760trr = 1.5 atm; P = 1.5 atm 0.14 N P ttal N = 0.168 atm n N 0.168atm 15.0 L = 0.08 ml N 0.0806L atm 7K K ml 0.08 ml N 8.0g N 1 ml N =.1 g N 15. Heat lss by U = heat gain by heavy water; vlume f cube = (cube edge) Mass f heavy water = 1.00 10 ml Heat gain by heavy water = 4.11J 1.11g ml = 1110 g 1110 g (8.5 5.5)C = 1.4 10 4 J Heat lss by U = 1.4 10 4 J = 0.117J mass (00.0 8.5)C, mass = 7.0 10 g U 7.0 10 g U 1cm 19.05g = 7 cm ; cube edge = (7 cm ) 1/ =. cm

1 CHAPTER 6 THERMOCHEMISTRY Marathn Prblems 16. Pathway I: Step 1: (5.00 ml,.00 atm, 15.0 L) (5.00 ml,.00 atm, 55.0 L) w = PΔV = (.00 atm)(55.0 L 15.0 L) = 10. L atm w = 10. L atm 101. J L atm 1kJ = 1. kj 1000J Step 1 is at cnstant pressure. The heat released/gained at cnstant pressure = q p = ΔH. Frm the prblem, ΔH = nc p ΔT fr an ideal gas. Using the ideal gas law, let s substitute fr ΔT. Δ(PV) = Δ(nRT) = nrδt fr a specific gas sample. S: ΔT = ΔH = q p = nc p ΔT = nc p Δ(PV) nr Δ(PV) nr CpΔ(PV) ; Nte: Δ(PV) = (P V P 1 V 1 ) R Fr an ideal mnatmic gas, C p = 5 R; substituting int the abve equatin: ΔH = Δ(PV) 5 R R 5 = Δ(PV) ΔH = q p = 5 Δ(PV) = 5 (.00 atm 55.0 L.00 atm 15.0 L) = 00. L atm ΔH = q p = 00. L atm 101. J L atm 1kJ = 0.4 kj 1000J ΔE = q + w = 0.4 kj 1. kj = 18. kj Nte: We culd have used ΔE = nc v ΔT t calculate the same answer (ΔE = 18. kj). Step : (5.00 ml,.00 atm, 55.0 L) (5.00 ml, 6.00 atm, 0.0 L) In this step, neither pressure nr vlume are cnstant. We will need t determine q in a different way. Hwever, it will always hld fr an ideal gas that ΔE = nc v ΔT and ΔH = nc p ΔT. Δ(PV) ΔE = nc v ΔT = n R nr ΔPV ΔE = (10. 165) L atm = 67.5 L atm (Carry an extra significant figure.) ΔE = 67.5 L atm 101. J 1kJ = 6.8 kj L atm 1000J

CHAPTER 6 THERMOCHEMISTRY 1 5 Δ(PV) ΔH = nc p ΔT = n R nr 5 ΔPV ΔH = 5 (10. 165) L atm = 11 L atm (Carry an extra significant figure.) w = PΔV = (6.00 atm)(0.0 55.0) L = 10. L atm w = 10. L atm 101. J L atm 1kJ = 1. kj 1000J ΔE = q + w, 6.8 kj = q + 1. kj, q = 8.1 kj Summary: Path I Step 1 Step Ttal Pathway II: q 0.4 kj 8.1 kj. kj w 1. kj 1. kj 9.1 kj ΔE 18. kj 6.8 kj 11.4 kj ΔH 0.4 kj 11 kj 19 kj Step : (5.00 ml,.00 atm, 15.0 L) (5.00 ml, 6.00 atm, 15.0 L) Step is at cnstant vlume. The heat released/gained at cnstant vlume = q v = ΔE. Δ(PV) ΔE = nc v ΔT = n R nr ΔPV ΔE = q v = Δ(PV) = (6.00 atm 15.0 L.00 atm 15.0 L) ΔE = q v = (90.0 45.0) L atm = 67.5 L atm ΔE = q v = 67.5 L atm 101. J L atm w = PΔV = 0 because ΔV = 0 1kJ = 6.84 kj 1000J ΔH = ΔE + Δ(PV) = 67.5 L atm + 45.0 L atm = 11.5 L atm = 11.40 kj Step 4: (5.00 ml, 6.00 atm, 15.0 L) (5.00 ml, 6.00 atm, 0.0 L) Step 4 is at cnstant pressure s q p = ΔH. ΔH = q p = nc p ΔT = 5 Δ(PV) R nr 5 ΔPV ΔH = 5 (10. 90.0) L atm = 75 L atm

14 CHAPTER 6 THERMOCHEMISTRY ΔH = q p = 75 L atm 101. J L atm 1kJ = 7.6 kj 1000J w = PΔV = (6.00 atm)(0.0 15.0) L = 0. L atm w = 0. L atm 101. J L atm 1kJ =.0 kj 1000J ΔE = q + w = 7.6 kj.0 kj = 4.6 kj Summary: Path II Step Step 4 Ttal q 6.84 kj 7.6 kj 14.4 kj w 0.0 kj.0 kj ΔE 6.84 kj 4.6 kj 11.4 kj ΔH 11.40 kj 7.6 kj 19.0 kj State functins are independent f the particular pathway taken between tw states; path functins are dependent n the particular pathway. In this prblem, the verall values f ΔH and ΔE fr the tw pathways are the same (see the tw summaries f results); hence ΔH and ΔE are state functins. Hwever, the verall values f q and w fr the tw pathways are different; hence q and w are path functins. x y/ 17. C x H y + x CO + y/ H O [x(9.5) + y/ (4)] ΔH C H x y = 044.5, (9.5)x 11y Δ 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 1.00atm MM, MM f CO 0.0806L atm /H O mixture = 9.1 g/ml 47K 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) Δ H H, ΔH C H C 8 8 = 104 kj/ml