CHAPTER 6 THERMOCHEMISTRY. Questions

Size: px
Start display at page:

Download "CHAPTER 6 THERMOCHEMISTRY. Questions"

Transcription

1 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 = 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 kj a. 1 ml Hg = 90.7 kj required ml Hg 18

2 CHAPTER 6 THERMOCHEMISTRY 18 b. 1 ml O 90.7 kj 1/ ml O = 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) = 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

3 184 CHAPTER 6 THERMOCHEMISTRY Mass = 5.5 z 1lb 16 z 1 kg = kg.05 lb Velcity = h mi 1 h 60 min 1 min 60 s 1760 yd mi 1 m yd = 45 m s 1 KE = mv 1 45 m = kg s = 150 J 4. KE = 5 1 mv 1 = 5 1 kg.0 10 cm 1 m g 1000 g =.0 10 s 100 cm J 5. KE = 1 mv 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 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 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 = = 1 kj 9 a. E = q + w =!47 kj + 88 kj = 41 kj b. E = 8! 47 = 5 kj c. E = = 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

4 CHAPTER 6 THERMOCHEMISTRY 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 J = 77 J b. w =!P V =!1.90 atm(.80 L! 8.0 L) = 10.5 L atm E = q + w = 50. J = 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 P 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 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 E = q + w,!10.5 J = 5.5 J + w, w =!155.0 J atm (V f! L), V f! = 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

5 186 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) = ml H O(l) 18.0 g ml 1cm = 18.1 cm = 18.1 ml g w =!P V =!1.00 atm ( L! 0.6 L) = 0.6 L atm E = q + w =!40.66 kj +.10 kj =!7.56 kj 101. J L atm = 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 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 ml Fe 165 kj 4 ml Fe =!1650 kj; 1650 kj f heat released b ml Fe O 165 kj ml Fe O =!86 kj; 86 kj f heat released

6 CHAPTER 6 THERMOCHEMISTRY 187 c g Fe 1 ml Fe 165 kj =!7.9 kj; 7.9 kj f heat released g 4 ml Fe d g Fe 1 ml Fe g = ml Fe;.00 g O 1 ml O.00 g = ml O 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 ml O 165 kj ml O 57 kj 44. a 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 L atm 98 K K ml 8 = ml H kj ml H ml H O =! kj; kj f heat released 45. Frm Example 6., q = J. Because the heat transfer prcess is nly 60.% efficient, the ttal energy required is J J = J. 60. J Mass C H 8 = J 1 ml C H J g CH ml C H 8 8 = g C H a g CH 4 1 ml CH4 891 kj =!55.5 kj g CH ml CH 4 4 b. n = PV = RT 740. atm L L atm 98 K K ml = 9.8 ml CH ml 891 kj ml =! kj

7 188 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 g (98 K - 7 K) = J b. Mlar heat capacity = 0.4 J g Ag ml Ag = 6 J C ml c. 150 J = 0.4 J m (15. C! 1.0 C), m = 150 = g Ag 0.4.

8 CHAPTER 6 THERMOCHEMISTRY s = specific heat capacity = q m T 1 J = = J/ECCg 5.00 g ( ) C Frm Table 6.1, the substance is slid aluminum. 5. s = 585 J 15.6 g ( ) 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 ! 09T f = 15T f , 4T f = , 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 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 = J mass ( ) C, mass = = 8 g 55. Heat lss by Al + heat lss by Fe = heat gain by water; keeping all quantities psitive t avid sign errr:

9 190 CHAPTER 6 THERMOCHEMISTRY 0.89 J 5.00 g Al (100.0 C T f ) J 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 (4.5)T f = 407T f T f = 9850, T f =.7 C 10. J 56. Heat released t water = 5.0 g H g H g methane 50. J g methane = J Heat gain by water = 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 = g (18. C 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 g (75.0 C C), s = 100 J 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 L ml/l = ml f bth AgNO and HCl are reacted. Thus 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 = g (.40.60) C = 0 J

10 CHAPTER 6 THERMOCHEMISTRY 191 Heat lss = 0 J; this is the heat evlved (exthermic reactin) when 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 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 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 = = 00.0 ml Heat gain = 4.18J 1.0 g 00.0 ml ml (1. 4.6) C = 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 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

11 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 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 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 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 g CaCl ml CaCl heat gained by slutin = J = T f 5.0 C = ( ) g (T f 5.0 C) 10 = 14. C, T f = 14. C C = 9. C ml HCl L L = ml HCl ml Ba(OH) 0.00 L = ml Ba(OH) L T react with all the HCl present, / = ml Ba(OH) is required. Because ml Ba(OH) is present, HCl is the limiting reactant ml HCl 118 kj ml HCl Heat gained by slutin = 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 g T

12 CHAPTER 6 THERMOCHEMISTRY a. Heat gain by calrimeter = heat lss by CH 4 = 6.79 g CH 4 1 ml CH 80 kj 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 =! 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 = g = kj g kj Heat gain = 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 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 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.

13 194 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 = 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

14 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 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 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).

15 196 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 CaSO 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 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

16 CHAPTER 6 THERMOCHEMISTRY 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 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 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)

17 198 CHAPTER 6 THERMOCHEMISTRY H = 4 kj 90. kj 6 ml + ml ml ml 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 = ( ) kj = ml 169 kj ml 84. C H 4 (g) + O (g) CO (g) + H O(l) H =! kj H =! kj = (!9.5) kj + (!85.8) kj! f, C H 4 H f, C H 4 H f, C H 4! kj =!158.6 kj! H, = 5.5 kj/ml Energy Cnsumptin and Surces 85. C(s) + H O(g) H (g) + CO(g) H = kj ( 4 kj) = 1 kj 86. CO(g) + H (g) CH OH(l) H = 9 kj ( 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 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

18 CHAPTER 6 THERMOCHEMISTRY 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 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 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 = 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) ml CH kj 891 kj 4.4 L CH4 = 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 = 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: J J = J 80.0 J Mass f C H = ml CH J J 6.04 g C ml C H H =.97 g C H

19 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 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 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 g f quinne. Because we are at cnstant vlume, q V = E. E cmb = 5.0 kj 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) 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 = 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

20 CHAPTER 6 THERMOCHEMISTRY 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 = 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.

21 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 = ( ,900 g) T, T = = 5.96 C g C Final temperature = = 0.0 C 105. HCl(aq) + NaOH(aq) H O(l) + NaCl(aq) H =!56 kj ml HCl L L = ml HCl L ml NaOH L = 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 ml NaOH 56 kj ml NaOH =!4. kj; 4. kj f heat is released 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 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 = g ml 6.7 J g (4.0! 0.0)EC = J

22 CHAPTER 6 THERMOCHEMISTRY 0 Because the slutin gained heat, the reactin is exthermic; q =! J fr the reactin. H = J 1.50 ml BaSO 5 4 =! 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 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 = g 6.4 kj/g = 7.90 kj = heat gain by water and calrimeter Heat gain = 7.90 kj = C kg 6.66 kj kg T + T C 7.90 = ( ) 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 g ml + ml = 4.41 g ml ml Fr kg f fuel: 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 g mass is 5 ml + 4 ml = kj. ml ml Fr kg f fuel: g 4594 kj g = 719 kj Thus we get mre energy per kilgram frm the N O 4 /N H CH mixture.

23 04 CHAPTER 6 THERMOCHEMISTRY / 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 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

24 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) = L atm 101. J L atm =! 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 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)( K) = 940 J =.94 kj Because pressure is cnstant: E = q p + w = H + w = 0.7 kj + (.94 kj) = 7.8 kj 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 = kj/h s m min h 5640 kj ml = kj/h Percent efficiency = energy used per hur ttal energy per hur 100 = kj 100 = 0.9% kj 40. kj h 119. Energy used in 8.0 hurs = 40. kwh = s 600 s = kj h 10. kj Energy frm the sun in 8.0 hurs = s m 60 s 60 min 8.0 h = kj/m min h Only 1% f the sunlight is cnverted int electricity: 0.1 ( kj/m ) area = kj, area = 7 m

25 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 qt 946 ml 1.4 g gallns = g f cncentrated nitric gal qt ml acid slutin g HNO g slutin = g HNO g slutin g HNO 1 ml 6.0 g 1 ml Na CO ml HNO g Na CO ml Na CO = g Na CO There are ( /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: ml HNO 69 kj g HNO =! 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 kcal 4.18 kj kcal = 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 = 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 ml O ) (0) = 1.5 ml. E = H!P V = H! nrt E = 85.8 kj! 1.50 ml J/K E = 85.8 kj!.7 kj = 8.1 kj ml 98 K 1 kj 1000 J

26 CHAPTER 6 THERMOCHEMISTRY 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 = J q = 1.00 ml J/ml = J q = /ECCg 18.0 g (100. 0)EC = J q 4 = 1.00 ml J/ml = J q 5 =.0 J/ECCg 18.0 g ( )EC = J q ttal = q 1 + q + q + q 4 + q 5 = 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 = 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 g (195 0) C

27 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 = = L atm 7 K K ml.50 atm L = = 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 / = 1.80; Because the actual mle rati is smaller than the required mle rati, O in the numeratr is limiting 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; atm 15.0 L = = 0.08 ml N L atm 7 K K ml P = χ = 1.5 atm 0.14 N P ttal N = 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 = ml 1.11g ml = 1110 g

28 CHAPTER 6 THERMOCHEMISTRY 09 Heat gain by heavy water = 4.11 J 1110 g ( )EC = J Heat lss by U = J = J mass ( )EC, mass = g U g U 1cm 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 = J/ CCg( g)( C)(1 kj/1000 J) H O rxn q rxn = kj (carrying extra significant figures) Because H = 189 kj/ml fr the decmpsitin reactin, and because nly kj f heat was released fr this reactin, kj (1 ml X/189 kj) = ml X was reacted. Mlar mass f X =.7 g X ml X = 7 g/ml Frm the prblem, 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 g 18.0 g 1.00 ml X = 7 g =.00 ml CO +.50 ml H O ml ml.0 g 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

29 10 CHAPTER 6 THERMOCHEMISTRY V = height area; area = πr ; V = (59.8 cm)(π)(8.00 cm) 1 L 1000 cm T = = 0.67 K = 1.0 L PV 747 1atm trr (1.0 L) 760 trr n ttal = = RT L atm (0.67 K) K ml = ml = ml CO + ml O + ml A Ml A = ml ttal 0.00 ml CO 0.05 ml O = ml A Because 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) O (g) 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 J/K ml 1. L atm = 150 J = 1.5 kj, w = 1.5 kj 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 H H ) ( H ) f, O f, N f kj = [( 9.5) kj +.5( 86) kj ] H, C H N O H =.5 kj/ml = kj/ml f 5 9, C H N O f 5 9

30 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 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 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 = ml Thus: 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

CHAPTER 6 THERMOCHEMISTRY. Questions

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

More information

188 CHAPTER 6 THERMOCHEMISTRY

188 CHAPTER 6 THERMOCHEMISTRY 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.

More information

CHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK

CHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK CHEM 1413 Chapter 6 Hmewrk Questins TEXTBOOK HOMEWORK 6.25 A 27.7-g sample f the radiatr clant ethylene glycl releases 688 J f heat. What was the initial temperature f the sample if the final temperature

More information

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C?

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C? NOTES: Thermchemistry Part 1 - Heat HEAT- TEMPERATURE - Thermchemistry: the study f energy (in the frm f heat) changes that accmpany physical & chemical changes heat flws frm high t lw (ht cl) endthermic

More information

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY Energy- the capacity t d wrk r t prduce heat 1 st Law f Thermdynamics: Law f Cnservatin f Energy- energy can be cnverted frm ne frm t anther but it can be neither

More information

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY CHEMICAL REACTIONS INVOLVE ENERGY The study energy and its transrmatins is knwn as thermdynamics. The discussin thermdynamics invlve the cncepts energy, wrk, and heat. Types Energy Ptential energy is stred

More information

Lecture 16 Thermodynamics II

Lecture 16 Thermodynamics II Lecture 16 Thermdynamics II Calrimetry Hess s Law Enthalpy r Frmatin Cpyright 2013, 2011, 2009, 2008 AP Chem Slutins. All rights reserved. Fur Methds fr Finding H 1) Calculate it using average bnd enthalpies

More information

Unit 14 Thermochemistry Notes

Unit 14 Thermochemistry Notes Name KEY Perid CRHS Academic Chemistry Unit 14 Thermchemistry Ntes Quiz Date Exam Date Lab Dates Ntes, Hmewrk, Exam Reviews and Their KEYS lcated n CRHS Academic Chemistry Website: https://cincchem.pbwrks.cm

More information

Chemistry 114 First Hour Exam

Chemistry 114 First Hour Exam Chemistry 114 First Hur Exam Please shw all wrk fr partial credit Name: (4 pints) 1. (12 pints) Espress is made by frcing very ht water under high pressure thrugh finely grund, cmpacted cffee. (Wikipedia)

More information

CHAPTER SIX THERMOCHEMISTRY. Questions

CHAPTER SIX THERMOCHEMISTRY. Questions CHAPTER SIX Questions 9. A coee-cup calorimeter is at constant (atmospheric) pressure. The heat released or gained at constant pressure is H. A bomb calorimeter is at constant volume. The heat released

More information

Chapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w

Chapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w Islated, Clsed and Open Systems 9.1 Energy as a Reactant r a Prduct 9.2 Transferring Heat and Ding Wrk 9.5 Heats f Reactin and Calrimetry 9.6 Hess s Law and Standard Heats f Reactin 9.7 Heats f Reactin

More information

Chapter 4 Thermodynamics and Equilibrium

Chapter 4 Thermodynamics and Equilibrium Chapter Thermdynamics and Equilibrium Refer t the fllwing figures fr Exercises 1-6. Each represents the energies f fur mlecules at a given instant, and the dtted lines represent the allwed energies. Assume

More information

Thermodynamics and Equilibrium

Thermodynamics and Equilibrium Thermdynamics and Equilibrium Thermdynamics Thermdynamics is the study f the relatinship between heat and ther frms f energy in a chemical r physical prcess. We intrduced the thermdynamic prperty f enthalpy,

More information

Thermochemistry. Thermochemistry

Thermochemistry. Thermochemistry Thermchemistry Petrucci, Harwd and Herring: Chapter 7 CHEM 1000A 3.0 Thermchemistry 1 Thermchemistry The study energy in chemical reactins A sub-discipline thermdynamics Thermdynamics studies the bulk

More information

Lecture 4. The First Law of Thermodynamics

Lecture 4. The First Law of Thermodynamics Lecture 4. The First Law f Thermdynamics THERMODYNAMICS: Basic Cncepts Thermdynamics: (frm the Greek therme, meaning "heat" and, dynamis, meaning "pwer") is the study f energy cnversin between heat and

More information

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics Chapters 9 and 35 Thermchemistry and Chemical Thermdynamics 1 Cpyright (c) 011 by Michael A. Janusa, PhD. All rights reserved. Thermchemistry Thermchemistry is the study f the energy effects that accmpany

More information

Chapter 17 Free Energy and Thermodynamics

Chapter 17 Free Energy and Thermodynamics Chemistry: A Mlecular Apprach, 1 st Ed. Nivald Tr Chapter 17 Free Energy and Thermdynamics Ry Kennedy Massachusetts Bay Cmmunity Cllege Wellesley Hills, MA 2008, Prentice Hall First Law f Thermdynamics

More information

CHEM 103 Calorimetry and Hess s Law

CHEM 103 Calorimetry and Hess s Law CHEM 103 Calrimetry and Hess s Law Lecture Ntes March 23, 2006 Prf. Sevian Annuncements Exam #2 is next Thursday, March 30 Study guide, practice exam, and practice exam answer key are already psted n the

More information

REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj

REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj Chemistry 102 ANSWER KEY REVIEW QUESTIONS Chapter 18 1. Calculate the heat reactin ( H ) in kj/ml r the reactin shwn belw, given the H values r each substance: NH (g) + F 2 (g) NF (g) + HF (g) H (kj/ml)

More information

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change?

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change? Name Chem 163 Sectin: Team Number: ALE 21. Gibbs Free Energy (Reference: 20.3 Silberberg 5 th editin) At what temperature des the spntaneity f a reactin change? The Mdel: The Definitin f Free Energy S

More information

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes.

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes. Edexcel IGCSE Chemistry Tpic 1: Principles f chemistry Chemical frmulae, equatins and calculatins Ntes 1.25 write wrd equatins and balanced chemical equatins (including state symbls): fr reactins studied

More information

AP Chemistry Assessment 2

AP Chemistry Assessment 2 AP Chemistry Assessment 2 DATE OF ADMINISTRATION: January 8 January 12 TOPICS COVERED: Fundatinal Tpics, Reactins, Gases, Thermchemistry, Atmic Structure, Peridicity, and Bnding. MULTIPLE CHOICE KEY AND

More information

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review Review Accrding t the nd law f Thermdynamics, a prcess is spntaneus if S universe = S system + S surrundings > 0 Even thugh S system

More information

How can standard heats of formation be used to calculate the heat of a reaction?

How can standard heats of formation be used to calculate the heat of a reaction? Answer Key ALE 28. ess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 4 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk neatly using dimensinal analysis

More information

A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1

A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1 A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1 (Nte: questins 1 t 14 are meant t be dne WITHOUT calculatrs!) 1.Which f the fllwing is prbably true fr a slid slute with a highly endthermic heat

More information

When a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q

When a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q Chemistry Ntes Lecture 15 [st] 3/6/09 IMPORTANT NOTES: -( We finished using the lecture slides frm lecture 14) -In class the challenge prblem was passed ut, it is due Tuesday at :00 P.M. SHARP, :01 is

More information

How can standard heats of formation be used to calculate the heat of a reaction?

How can standard heats of formation be used to calculate the heat of a reaction? Name Chem 161, Sectin: Grup Number: ALE 28. Hess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 5 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk

More information

N 2 (g) + 3H 2 (g) 2NH 3 (g) o Three mole ratios can be derived from the balanced equation above: Example: Li(s) + O 2 (g) Li 2 O(s)

N 2 (g) + 3H 2 (g) 2NH 3 (g) o Three mole ratios can be derived from the balanced equation above: Example: Li(s) + O 2 (g) Li 2 O(s) Chapter 9 - Stichimetry Sectin 9.1 Intrductin t Stichimetry Types f Stichimetry Prblems Given is in mles and unknwn is in mles. Given is in mles and unknwn is in mass (grams). Given is in mass and unknwn

More information

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses Learning Objectives 17.1: Spntaneus Prcesses Cmparing and Cntrasting the hree Laws f hermdynamics (1 st Law: Chap. 5; 2 nd & 3

More information

CHM 152 Practice Final

CHM 152 Practice Final CM 152 Practice Final 1. Of the fllwing, the ne that wuld have the greatest entrpy (if cmpared at the same temperature) is, [a] 2 O (s) [b] 2 O (l) [c] 2 O (g) [d] All wuld have the same entrpy at the

More information

Thermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P

Thermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P Thermchemistry The study energy changes that ccur during chemical : at cnstant vlume ΔU = q V n at cnstant pressure = q P nly wrk Fr practical reasns mst measurements are made at cnstant, s thermchemistry

More information

Chemistry 1A Fall 2000

Chemistry 1A Fall 2000 Chemistry 1A Fall 2000 Midterm Exam III, versin B Nvember 14, 2000 (Clsed bk, 90 minutes, 155 pints) Name: SID: Sectin Number: T.A. Name: Exam infrmatin, extra directins, and useful hints t maximize yur

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25 CHAPTER 17 1. Read Chapter 17, sectins 1,2,3. End f Chapter prblems: 25 2. Suppse yu are playing a game that uses tw dice. If yu cunt the dts n the dice, yu culd have anywhere frm 2 t 12. The ways f prducing

More information

Chem 75 February 16, 2017 Exam 2 Solutions

Chem 75 February 16, 2017 Exam 2 Solutions 1. (6 + 6 pints) Tw quick questins: (a) The Handbk f Chemistry and Physics tells us, crrectly, that CCl 4 bils nrmally at 76.7 C, but its mlar enthalpy f vaprizatin is listed in ne place as 34.6 kj ml

More information

Chem 111 Summer 2013 Key III Whelan

Chem 111 Summer 2013 Key III Whelan Chem 111 Summer 2013 Key III Whelan Questin 1 6 Pints Classify each f the fllwing mlecules as plar r nnplar? a) NO + : c) CH 2 Cl 2 : b) XeF 4 : Questin 2 The hypthetical mlecule PY 3 Z 2 has the general

More information

4 Fe + 3 O 2 2 Fe 2 O 3

4 Fe + 3 O 2 2 Fe 2 O 3 UNIT 7: STOICHIOMETRY NOTES (chapter 9) INTRO TO STOICHIOMETRY Reactin Stichimetry: Stichimetry is simply a way t shw f smething this is. Relatinship between a given and an unknwn: GIVEN UNKNOWN Type 1

More information

Semester 2 AP Chemistry Unit 12

Semester 2 AP Chemistry Unit 12 Cmmn In Effect and Buffers PwerPint The cmmn in effect The shift in equilibrium caused by the additin f a cmpund having an in in cmmn with the disslved substance The presence f the excess ins frm the disslved

More information

A Chemical Reaction occurs when the of a substance changes.

A Chemical Reaction occurs when the of a substance changes. Perid: Unit 8 Chemical Reactin- Guided Ntes Chemical Reactins A Chemical Reactin ccurs when the f a substance changes. Chemical Reactin: ne r mre substances are changed int ne r mre new substances by the

More information

Chapter 9 Chemical Reactions NOTES

Chapter 9 Chemical Reactions NOTES Chapter 9 Chemical Reactins NOTES Chemical Reactins Chemical reactin: Chemical change 4 Indicatrs f Chemical Change: (1) (2) (3) (4) Cnsist f reactants (starting materials) and prducts (substances frmed)

More information

CHE 105 EXAMINATION III November 11, 2010

CHE 105 EXAMINATION III November 11, 2010 CHE 105 EXAMINATION III Nvember 11, 2010 University f Kentucky Department f Chemistry READ THESE DIRECTIONS CAREFULLY BEFORE STARTING THE EXAMINATION! It is extremely imprtant that yu fill in the answer

More information

Entropy, Free Energy, and Equilibrium

Entropy, Free Energy, and Equilibrium Nv. 26 Chapter 19 Chemical Thermdynamics Entrpy, Free Energy, and Equilibrium Nv. 26 Spntaneus Physical and Chemical Prcesses Thermdynamics: cncerned with the questin: can a reactin ccur? A waterfall runs

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Recitation 06. n total = P total V/RT = (0.425 atm * 10.5 L) / ( L atm mol -1 K -1 * 338 K) = mol

Recitation 06. n total = P total V/RT = (0.425 atm * 10.5 L) / ( L atm mol -1 K -1 * 338 K) = mol Recitatin 06 Mixture f Ideal Gases 1. Chapter 5: Exercise: 69 The partial pressure f CH 4 (g) is 0.175 atm and that f O 2 (g) is 0.250 atm in a mixture f the tw gases. a. What is the mle fractin f each

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t eep this site up and bring yu even mre cntent cnsider dnating via the lin n ur site. Still having truble understanding the material? Chec ut ur Tutring

More information

Review Material for Exam #2

Review Material for Exam #2 Review Material fr Eam # 1. a. Calculate the mlarity f a slutin made with 184.6 mg sample f ptassium dichrmate disslved in enugh water t give 500.0 ml f slutin. 1 g 184.6 mg K Cr O 1 mle K Cr O 7 7 1000

More information

Thermochemistry Heats of Reaction

Thermochemistry Heats of Reaction hermchemistry Heats f Reactin aa + bb cc + dd hermchemical Semantics q V = Heat f Rxn at [V] = U = Energy (change) f Rxn q P = Heat f Rxn at [P] = H = Enthalpy (change) f Rxn Exthermic rxns: q < 0 Endthermic

More information

Unit 11 Solutions- Guided Notes. What are alloys? What is the difference between heterogeneous and homogeneous mixtures?

Unit 11 Solutions- Guided Notes. What are alloys? What is the difference between heterogeneous and homogeneous mixtures? Name: Perid: Unit 11 Slutins- Guided Ntes Mixtures: What is a mixture and give examples? What is a pure substance? What are allys? What is the difference between hetergeneus and hmgeneus mixtures? Slutins:

More information

NUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command

NUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command NUPOC SUDY GUIDE ANSWER KEY Navy Recruiting Cmmand CHEMISRY. ph represents the cncentratin f H ins in a slutin, [H ]. ph is a lg scale base and equal t lg[h ]. A ph f 7 is a neutral slutin. PH < 7 is acidic

More information

Chemistry/ Biotechnology Reference Sheets

Chemistry/ Biotechnology Reference Sheets Cmmn Metric Prefixes: Giga (G) = 1,000,000,000 = Kil (k) = 1,000 = Deci (d) =.1 = Milli (m) =.001 = Nan (n) =.000000001 = 9 6 1 10 Mega (M) = 1,000,000 = 1 10 0 1 10 Basic unit = meter, gram, liter, secnd

More information

33. a. Heat is absorbed from the water (it gets colder) as KBr dissolves, so this is an endothermic process.

33. a. Heat is absorbed from the water (it gets colder) as KBr dissolves, so this is an endothermic process. 31. This is an endothermic reaction so heat must be absorbed in order to convert reactants into products. The high temperature environment of internal combustion engines provides the heat. 33. a. Heat

More information

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia:

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia: University Chemistry Quiz 3 2015/04/21 1. (10%) Cnsider the xidatin f ammnia: 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(l) (a) Calculate the ΔG fr the reactin. (b) If this reactin were used in a fuel cell,

More information

Hess Law - Enthalpy of Formation of Solid NH 4 Cl

Hess Law - Enthalpy of Formation of Solid NH 4 Cl Hess Law - Enthalpy f Frmatin f Slid NH 4 l NAME: OURSE: PERIOD: Prelab 1. Write and balance net inic equatins fr Reactin 2 and Reactin 3. Reactin 2: Reactin 3: 2. Shw that the alebraic sum f the balanced

More information

Chem 116 POGIL Worksheet - Week 3 - Solutions Intermolecular Forces, Liquids, Solids, and Solutions

Chem 116 POGIL Worksheet - Week 3 - Solutions Intermolecular Forces, Liquids, Solids, and Solutions Chem 116 POGIL Wrksheet - Week 3 - Slutins Intermlecular Frces, Liquids, Slids, and Slutins Key Questins 1. Is the average kinetic energy f mlecules greater r lesser than the energy f intermlecular frces

More information

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review)

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review) CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUM Part One: Heat Changes and Thermchemistry This aspect f Thermdynamics was dealt with in Chapter 6. (Review) A. Statement f First Law. (Sectin 18.1) 1. U ttal

More information

15.0 g Cr = 21.9 g Cr O g Cr 4 mol Cr mol Cr O

15.0 g Cr = 21.9 g Cr O g Cr 4 mol Cr mol Cr O WYSE Academic Challenge Sectinal Chemistry Exam 2008 SOLUTION SET 1. Crrect answer: B. Use PV = nrt t get: PV = nrt 2. Crrect answer: A. (2.18 atm)(25.0 L) = n(0.08206 L atm/ml K)(23+273) n = 2.24 ml Assume

More information

Matter Content from State Frameworks and Other State Documents

Matter Content from State Frameworks and Other State Documents Atms and Mlecules Mlecules are made f smaller entities (atms) which are bnded tgether. Therefre mlecules are divisible. Miscnceptin: Element and atm are synnyms. Prper cnceptin: Elements are atms with

More information

CHAPTER PRACTICE PROBLEMS CHEMISTRY

CHAPTER PRACTICE PROBLEMS CHEMISTRY Chemical Kinetics Name: Batch: Date: Rate f reactin. 4NH 3 (g) + 5O (g) à 4NO (g) + 6 H O (g) If the rate f frmatin f NO is 3.6 0 3 ml L s, calculate (i) the rate f disappearance f NH 3 (ii) rate f frmatin

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

BIT Chapters = =

BIT Chapters = = BIT Chapters 17-0 1. K w = [H + ][OH ] = 9.5 10 14 [H + ] = [OH ] =.1 10 7 ph = 6.51 The slutin is neither acidic nr basic because the cncentratin f the hydrnium in equals the cncentratin f the hydride

More information

General Chemistry II, Unit II: Study Guide (part 1)

General Chemistry II, Unit II: Study Guide (part 1) General Chemistry II, Unit II: Study Guide (part 1) CDS Chapter 21: Reactin Equilibrium in the Gas Phase General Chemistry II Unit II Part 1 1 Intrductin Sme chemical reactins have a significant amunt

More information

GOAL... ability to predict

GOAL... ability to predict THERMODYNAMICS Chapter 18, 11.5 Study f changes in energy and transfers f energy (system < = > surrundings) that accmpany chemical and physical prcesses. GOAL............................. ability t predict

More information

CHEM 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions?

CHEM 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions? 1 CHEM 1032 FALL 2017 Practice Exam 4 1. Which f the fllwing reactins is spntaneus under nrmal and standard cnditins? A. 2 NaCl(aq) 2 Na(s) + Cl2(g) B. CaBr2(aq) + 2 H2O(aq) Ca(OH)2(aq) + 2 HBr(aq) C.

More information

3. Review on Energy Balances

3. Review on Energy Balances 3. Review n Energy Balances Objectives After cmpleting this chapter, students shuld be able t recall the law f cnservatin f energy recall hw t calculate specific enthalpy recall the meaning f heat f frmatin

More information

A. Lattice Enthalpies Combining equations for the first ionization energy and first electron affinity:

A. Lattice Enthalpies Combining equations for the first ionization energy and first electron affinity: [15.1B Energy Cycles Lattice Enthalpy] pg. 1 f 5 CURRICULUM Representative equatins (eg M+(g) M+(aq)) can be used fr enthalpy/energy f hydratin, inizatin, atmizatin, electrn affinity, lattice, cvalent

More information

General Chemistry II, Unit I: Study Guide (part I)

General Chemistry II, Unit I: Study Guide (part I) 1 General Chemistry II, Unit I: Study Guide (part I) CDS Chapter 14: Physical Prperties f Gases Observatin 1: Pressure- Vlume Measurements n Gases The spring f air is measured as pressure, defined as the

More information

Tuesday, 5:10PM FORM A March 18,

Tuesday, 5:10PM FORM A March 18, Name Chemistry 153-080 (3150:153-080) EXAM II Multiple-Chice Prtin Instructins: Tuesday, 5:10PM FORM A March 18, 2003 120 1. Each student is respnsible fr fllwing instructins. Read this page carefully.

More information

Spontaneous Processes, Entropy and the Second Law of Thermodynamics

Spontaneous Processes, Entropy and the Second Law of Thermodynamics Chemical Thermdynamics Spntaneus Prcesses, Entrpy and the Secnd Law f Thermdynamics Review Reactin Rates, Energies, and Equilibrium Althugh a reactin may be energetically favrable (i.e. prducts have lwer

More information

CHEM 1001 Problem Set #3: Entropy and Free Energy

CHEM 1001 Problem Set #3: Entropy and Free Energy CHEM 1001 Prblem Set #3: Entry and Free Energy 19.7 (a) Negative; A liquid (mderate entry) cmbines with a slid t frm anther slid. (b)psitive; One mle f high entry gas frms where n gas was resent befre.

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Heat is energy and is measured in joules (J) or kilojoules (kj). The symbol for heat is H.

Heat is energy and is measured in joules (J) or kilojoules (kj). The symbol for heat is H. Causes f Change Calrimetry Hw Des Energy Affect Change? Heat vs. Temerature HEAT TEMPERATURE Definitin: Deends n: Examles: Heat is energy and is measured in jules (J) r kiljules (kj). The symbl fr heat

More information

Solutions to the Extra Problems for Chapter 14

Solutions to the Extra Problems for Chapter 14 Slutins t the Extra Prblems r Chapter 1 1. The H -670. T use bnd energies, we have t igure ut what bnds are being brken and what bnds are being made, s we need t make Lewis structures r everything: + +

More information

Entropy and Gibbs energy

Entropy and Gibbs energy 14 Entrpy and Gibbs energy Answers t wrked examples WE 14.1 Predicting the sign f an entrpy change (n p. 658 in Chemistry 3 ) What will be the sign f the value f S fr: (a) crystallizatin f salt frm a slutin;

More information

Chapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry

Chapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry Chapter 19 lectrchemistry Part I Dr. Al Saadi 1 lectrchemistry What is electrchemistry? It is a branch f chemistry that studies chemical reactins called redx reactins which invlve electrn transfer. 19.1

More information

" 1 = # $H vap. Chapter 3 Problems

 1 = # $H vap. Chapter 3 Problems Chapter 3 rblems rblem At 1 atmsphere pure Ge melts at 1232 K and bils at 298 K. he triple pint ccurs at =8.4x1-8 atm. Estimate the heat f vaprizatin f Ge. he heat f vaprizatin is estimated frm the Clausius

More information

More Tutorial at

More Tutorial at Answer each questin in the space prvided; use back f page if extra space is needed. Answer questins s the grader can READILY understand yur wrk; nly wrk n the exam sheet will be cnsidered. Write answers,

More information

In the spaces provided, explain the meanings of the following terms. You may use an equation or diagram where appropriate.

In the spaces provided, explain the meanings of the following terms. You may use an equation or diagram where appropriate. CEM1405 2007-J-2 June 2007 In the spaces prvided, explain the meanings f the fllwing terms. Yu may use an equatin r diagram where apprpriate. 5 (a) hydrgen bnding An unusually strng diple-diple interactin

More information

Unit 9: The Mole- Guided Notes What is a Mole?

Unit 9: The Mole- Guided Notes What is a Mole? Unit 9: The Mle- Guided Ntes What is a Mle? A mle is a name fr a specific f things Similar t a r a One mle is equal t 602 602,000,000,000,000,000,000,000 That s 602 with zers A mle is NOT an abbreviatin

More information

Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra

Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra Chem 115 POGIL Wrksheet - Week 8 Thermchemistry (Cntinued), Electrmagnetic Radiatin, and Line Spectra Why? As we saw last week, enthalpy and internal energy are state functins, which means that the sum

More information

Work and Heat Definitions

Work and Heat Definitions Wrk and eat Deinitins FL- Surrundings: Everything utside system + q -q + System: he part S the rld e are bserving. Wrk, : transer energy as a result unbalanced rces - eat, q: transer energy resulting rm

More information

SPONTANEITY, ENTROPY, AND FREE ENERGY

SPONTANEITY, ENTROPY, AND FREE ENERGY CHAER 7 SONANEIY, ENROY, AND FREE ENERGY Questins. Living rganisms need an external surce f energy t carry ut these prcesses. Green plants use the energy frm sunlight t prduce glucse frm carbn dixide and

More information

Thermodynamics Partial Outline of Topics

Thermodynamics Partial Outline of Topics Thermdynamics Partial Outline f Tpics I. The secnd law f thermdynamics addresses the issue f spntaneity and invlves a functin called entrpy (S): If a prcess is spntaneus, then Suniverse > 0 (2 nd Law!)

More information

Advanced Chemistry Practice Problems

Advanced Chemistry Practice Problems Advanced Chemistry Practice Prblems Thermdynamics: Gibbs Free Energy 1. Questin: Is the reactin spntaneus when ΔG < 0? ΔG > 0? Answer: The reactin is spntaneus when ΔG < 0. 2. Questin: Fr a reactin with

More information

Thermodynamics I. Prep Session

Thermodynamics I. Prep Session Thermodynamics I Prep Session Dr. John I. Gelder Department of Chemistry Oklahoma State University Stillwater, OK 74078 john.gelder@okstate.edu http://intro.chem.okstate.edu 12/5/09 1 Thermo I Prep Session

More information

Acids and Bases Lesson 3

Acids and Bases Lesson 3 Acids and Bases Lessn 3 The ph f a slutin is defined as the negative lgarithm, t the base ten, f the hydrnium in cncentratin. In a neutral slutin at 25 C, the hydrnium in and the hydrxide in cncentratins

More information

Three Definitions of Acids/Bases Type Acid Base Problems with it Arrhenius Bronsted-Lowry Lewis. NH 3(aq) + H 2O(l)

Three Definitions of Acids/Bases Type Acid Base Problems with it Arrhenius Bronsted-Lowry Lewis. NH 3(aq) + H 2O(l) CP NT Ch 19: Acid and Bases An Intrductin Prperties f Acids 1. taste 2. Can prduce H + ( ) ins ( ) 3. Change the clr f litmus frm t 4. Reacts with such as Zn and Mg t prduce gas. Ba(s) + H 2SO 4 BaSO 4(aq)

More information

( ) kt. Solution. From kinetic theory (visualized in Figure 1Q9-1), 1 2 rms = 2. = 1368 m/s

( ) kt. Solution. From kinetic theory (visualized in Figure 1Q9-1), 1 2 rms = 2. = 1368 m/s .9 Kinetic Mlecular Thery Calculate the effective (rms) speeds f the He and Ne atms in the He-Ne gas laser tube at rm temperature (300 K). Slutin T find the rt mean square velcity (v rms ) f He atms at

More information

10/23/10. Thermodynamics and Kinetics. Chemical Hand Warmers

10/23/10. Thermodynamics and Kinetics. Chemical Hand Warmers 10/23/10 CHAPTER 6 Thermochemistry 6-1 Chemical Hand Warmers Most hand warmers work by using the heat released from the slow oxidation of iron 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) The amount your hand temperature

More information

/ / Chemistry. Chapter 1 Chemical Foundations

/ / Chemistry. Chapter 1 Chemical Foundations Name Chapter 1 Chemical Fundatins Advanced Chemistry / / Metric Cnversins All measurements in chemistry are made using the metric system. In using the metric system yu must be able t cnvert between ne

More information

Chapter One. Matter and Energy - Chemistry the study of matter and its changes the "central science" Natural Laws

Chapter One. Matter and Energy - Chemistry the study of matter and its changes the central science Natural Laws Chapter One Matter and Measurement http://www.chemistry.armstrng.edu/ nivens/curse_list.htm OWL HOMEWORK REQUIRED!!! Matter and Energy - Chemistry the study f matter and its changes the "central science"

More information

Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals of Diffusion

Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals of Diffusion Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals f Diffusin Diffusin: Transprt in a slid, liquid, r gas driven by a cncentratin gradient (r, in the case f mass transprt, a chemical ptential

More information

CHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics

CHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics CHEM 116 Electrchemistry at Nn-Standard Cnditins, and Intr t Thermdynamics Imprtant annuncement: If yu brrwed a clicker frm me this semester, return it t me at the end f next lecture r at the final exam

More information

5.0 minutes. The temperature rose from

5.0 minutes. The temperature rose from QUESTION 1 The table belw describes features cmmn t a slutin calrimeter and a bmb calrimeter as well as differences. Features f Bth the Slutin and Bmb alrimeters W X Features f the Bmb alrimeter Only Y

More information

Answer Key, Problem Set 8b (full)

Answer Key, Problem Set 8b (full) Chemistry 11 Mines, Fall 017 Answer Key, Prblem Set 8b (full) 1. NT1;. NT; 3. 6.58 (with extra parts added); 4. NT3; 5. 6. & 6.3; 6. 6.93; 7. 6.99; 8. 6.75 (i.e., determine the H fr the thermchemical equatin);

More information

GASES. PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2. Pressure & Boyle s Law Temperature & Charles s Law Avogadro s Law IDEAL GAS LAW

GASES. PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2. Pressure & Boyle s Law Temperature & Charles s Law Avogadro s Law IDEAL GAS LAW GASES Pressure & Byle s Law Temperature & Charles s Law Avgadr s Law IDEAL GAS LAW PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2 Earth s atmsphere: 78% N 2 21% O 2 sme Ar, CO 2 Sme Cmmn Gasses Frmula Name

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

CHM112 Lab Graphing with Excel Grading Rubric

CHM112 Lab Graphing with Excel Grading Rubric Name CHM112 Lab Graphing with Excel Grading Rubric Criteria Pints pssible Pints earned Graphs crrectly pltted and adhere t all guidelines (including descriptive title, prperly frmatted axes, trendline

More information

In the half reaction I 2 2 I the iodine is (a) reduced (b) oxidized (c) neither of the above

In the half reaction I 2 2 I the iodine is (a) reduced (b) oxidized (c) neither of the above 6.3-110 In the half reactin I 2 2 I the idine is (a) reduced (b) xidized (c) neither f the abve 6.3-120 Vitamin C is an "antixidant". This is because it (a) xidizes readily (b) is an xidizing agent (c)

More information

Student Exploration: Cell Energy Cycle

Student Exploration: Cell Energy Cycle Name: Date: Student Explratin: Cell Energy Cycle Vcabulary: aerbic respiratin, anaerbic respiratin, ATP, cellular respiratin, chemical energy, chlrphyll, chlrplast, cytplasm, glucse, glyclysis, mitchndria,

More information