Chapter 11 REVIEW. Part 1

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1 produces soe hydroelectricity. This involves energy conversions fro potential energy of water (above the da) to kinetic energy (falling water and spinning turbine) to electrical energy. Coal accounts for ~66% of the generated electricity in Alberta, gas for ~23%, and hydro for ~10%. 3. Benefits of Coal plentiful easy to convert to electrical energy (proven technology) inexpensive Risks of Coal causes pollution (produces NO x and SO 2 ) contributes to global waring coal ining can result in environental daage 4. solar energy cheical energy (coal) theral energy (H 2 O) kinetic energy (stea oving the turbine that turns the generator) electrical energy light and heat 5. (a) If the reaction occurs in a particular ediu (e.g., water), the change in teperature of that ediu can be easured. This change in teperature is a reflection of the quantity of energy absorbed or released by the reaction in the ediu, Q = ct. (b) The energy released or absorbed by a cheical reaction is counicated by its enthalpy change, r H, for a specified cheical equation, or by its olar enthalpy change, r H, for a given substance reacting in a specified reaction. Chapter 11 REVIEW Part 1 (Page 519) 1. B 2. 1, 4, 3, 2 3. C D C 8. D 9. 1, 3, , 4, , 4, 2, , 2, 1, , 2, 4, [Note error in text: should be labelled +] Solutions 4.19 J 4. Q ct 1.50 kg ( ) C 585 kj g C kj 6. ch 98.9 kj/ol SO 2ol 2 rh ch nfph nfrh and ch n 12. ch : kj/ol; kj/ol; kj/ol; kj/ol CH CH CH CHOH Unit 6 Solutions Manual Copyright 2007 Thoson Nelson

2 14. C 2 H 6 (g) C 2 H 4 (g) + H 2 (g) r H =? 52.4 kj 0 kj rh 1ol C2H4 1olH2 1olC2H4 1olH kj 1olC2H6 1olC2H kj Part 2 (Pages ) 15. The sun s energy was initially trapped by photosynthesis in green plants to produce organic copounds. The plant atter (either eaten by anials or left in the plant) eventually was converted by high teperatures and pressures into hydrocarbons. 16. The enthalpy change, r H, refers to the total energy change for a cheical syste or substance undergoing change, usually described by a cheical reaction equation. The olar enthalpy, r H, change refers to the energy change per ole of the substance undergoing the change indicated; e.g., cobustion and foration. 17. (a) Q = ct = 2.0 kg 4.19 J/(g C) 5.3 C = 44 kj 44 kj (b) Q/ = = 36 kj/g 1.25 g 18. (a) Q = ct = 2.23 kg 4.19 J/(g C) 8.04 C = 75.1 kj (b) Because cobustion reactions are exotheric and r H = Q, ch 75.1 kj 75.1 kj (c) ch CH 1ol g g 2.87 MJ/ol ch 2.87 MJ/ol The sybol, value, and sign would be. CH The H represents the enthalpy change; the subscript c specifies a cobustion reaction; and the subscript indicates that this is a olar quantity of the substance specified under the sybol. The value is obtained fro calorietry and the negative sign indicates that this is an exotheric reaction, on the basis of the generalization that all cobustion reactions are exotheric. 19. Proble What is the olar enthalpy of neutralization for ethanoic acid? Analysis Qct 4.19 J 100 g ( ) C g C 2.22 kj 4 10 Copyright 2007 Thoson Nelson Unit 6 Solutions Manual 449

3 2.22 kj nh ethanoicacid 1.00 ol L 1 L 44.4 kj/ol According to the evidence, the olar enthalpy of neutralization for ethanoic acid is 44.4 kj/ol. 20. (a) Q = ct = 100 kg 4.19 J/(g C) ( ) C = 21.8 MJ (b) CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) kj kj resh 1ol CO2 2olH2O 1 olco2 1 olh2o 74.6 kj 0 kj 1olCH4 2olO2 1olCH4 1olO kj 21.8MJ CH kj 1ol 1 ol 16.05g 0.436kg (c) Q = ct 32.5 MJ = 4.19 J/(g C) ( ) C = Mg or 234 kg v = 234 L 21. Coparing Photosynthesis, Cellular Respiration, and Hydrocarbon Cobustion Reaction Reactants Products Endotheric/ Exotheric Enthalpy change +/ photosynthesis CO 2 + H 2 O C 6 H 12 O 6 + O 2 endotheric + cellular C 6 H 12 O 6 + O 2 CO 2 + H 2 O exotheric respiration cobustion hydrocarbon + O 2 CO 2 + H 2 O exotheric 22. C 2 H 2 (g) O 2(g) 2 CO 2 (g) + H 2 O(g) kj kj ch 2 ol CO2 1 olh2o 1olCO2 1olH2O 227.4kJ 5 0kJ 1olC2H2 olo2 1olC2H2 2 1olO 2 ( kJ) ( kj) [ For use in the graph below. ] kJ kj H kj/ ol 1ol c CH 2 2 (i) c H = kj/ol C 2 H 2 (ii) C 2 H 2 (g) O 2(g) 2 CO 2 (g) + H 2 O(g) c H = kj 450 Unit 6 Solutions Manual Copyright 2007 Thoson Nelson

4 (iii) C 2 H 2 (g) O 2(g) 2 CO 2 (g) + H 2 O(g) kj (iv) 23. The olar enthalpies of foration ethod can be used to deterine the enthalpy change for a particular reaction. The standard olar enthalpies of foration for each of the entities in the desired reaction can be obtained fro a table of standard olar enthalpies of foration. These values can then be used in the equation: r H = n fp H n fr H to find the enthalpy change for the reaction. Alternatively, a version of Hess law, which cobines known equations for which enthalpy changes have been established, can be used. In this case, the equations are anipulated and added to produce the desired net equation. The su of the enthalpy changes for the individual equations, which result in the desired equation, will yield the enthalpy change for the desired process. neth rh 24. C 2 H 4 (g) + Cl 2 (g) CH 2 ClCH 2 Cl(l) H = kj CH 2 ClCH 2 Cl(l) C 2 H 3 Cl(g) + HCl(g) H = kj C 2 H 4 (g) + Cl 2 (g) C 2 H 3 Cl(g) + HCl(g) H = kj 25. C 16 H 34 (l) + H 2 (g) 2 C 8 H 18 (l) r H =? C 16 H 34 (l) O 2(g) 16 CO 2 (g) + 17 H 2 O(l) r H = kj 16 CO 2 (g) + 18 H 2 O(l) 2 C 8 H 18 (l) + 25 O 2 (g) r H = kj H 2 (g) O 2(g) H 2 O(l) net: C 16 H 34 (l) + H 2 (g) 2 C 8 H 18 (l) 26. (a) CuS(s) O 2(g) CuO(s) + SO 2 (g) kj kj ch 1ol CuO 1olSO2 1olCuO 1olSO kj 3 0 kj 1 olcus olo2 1olCuS 2 1olO 2 ( 454.1kJ) ( 53.1kJ) [ For use in the graph below.] 401.0kJ kj rh kj/ol CuS 1ol r H = 286 kj r H = 69 kj Copyright 2007 Thoson Nelson Unit 6 Solutions Manual 451

5 (b) 27. (a) C 8 H 18 (l) + H 2 (g) C 5 H 12 (l) + C 3 H 8 (g) kj kj rh 1ol C5H12 1olC3H8 1olC5H12 1olC3H kj 0 kj 1 olc8h18 1 olh2 1olC8H18 1olH kj (b) C 2 H 4 (g) + C 6 H 6 (l) C 6 H 5 CHCH 2 + H 2 (g) kj 0 kj rh 1ol C6H5CHCH2 1olH2 1 olc6h5chch2 1 olh kj 49.1 kj 1olC2H4 1olC6H6 1 olc2h4 1 olc6h 6 2.3kJ 28. Proble What is the enthalpy change for the reaction? Ca(s) + ½ O 2 (g) CaO(s) Prediction According to the table of olar enthalpies of foration, the enthalpy change is kj because the cheical equation represents the foration of one ole of calciu oxide. Analysis Reaction #1 Ca(s) + HCl(aq) CaCl 2 (aq) + H 2 (g) Q ct (calorieter) 100 g 4.19J g C ( ) C 5.53 kj 5.53 kj rh Ca 1ol 0.52 g g 426 kj/ol According to the evidence, the enthalpy change for the reaction, based on one ole of calciu, is: Ca(s) + HCl(aq) CaCl 2 (aq) + H 2 (g) r H = 426 kj 452 Unit 6 Solutions Manual Copyright 2007 Thoson Nelson

6 Reaction #2 CaO(s) + 2 HCl(aq) CaCl 2 (aq) + H 2 O(l) Q ct (calorieter) 100 g 4.19J g C ( ) C = 2.89 kj 2.89 kj rh CaO 1ol 1.47 g g 110 kj/ol According to the evidence, the enthalpy change for the reaction, based on one ole of calciu oxide, is CaO(s) + 2 HCl(aq) CaCl 2 (aq) + H 2 O(l) r H = 110 kj Reaction #3 According to the table of olar enthalpies of foration, H 2 (g) O 2(g) H 2 O(l) Applying Hess law, Ca(s) + 2 HCl(aq) CaCl2(aq) + H2(g) CaCl 2 (aq) + H 2 O(l) CaO(s) + 2 HCl(aq) H 2 (g) O 2(g) H 2 O(l) net: Ca(s) O 2(g) CaO(s) f H = kj rh = 426 kj r H = +110 kj f H = kj f H = 602 kj According to the evidence collected and Hess law, the enthalpy change for the given reaction is 602 kj. 29. (a) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) kj kj ch 3ol CO2 4olH2O 1 olco2 1 olh2o kj 0 kj 1olC3H8 5olO2 1olC3H8 1olO kJ kj ch kj/ol CH 1ol 3 8 (b) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) kj kj ch 3ol CO2 4olH2O 1 olco2 1 olh2o kj 0 kj 1olC3H8 5olO2 1olC3H8 1olO kJ kJ H kJ/ol 1ol c CH 3 8 Copyright 2007 Thoson Nelson Unit 6 Solutions Manual 453

7 (c) 2 C 3 H 8 (g) + 8 O 2 (g) C(s) + 2 CO(g) + 3 CO 2 (g) + 8 H 2 O(g) 0 kj kj kj rh 1 ol C + 2 ol CO + 3 ol CO 2 1 ol C 1 ol CO 1 ol CO kj kj 0 kj 8 ol H2O 2 ol C3H ol O2 1 ol H2O 1 ol C3H8 1 ol O kJ o kj rh = = kj/ol CH 2 ol 3 8 (d) incoplete cobustion, cobustion to produce water vapour, cobustion to produce liquid water (e) The ost efficient type of furnace is a condensing high-efficiency furnace. One way in which these devices differ fro conventional furnaces is that they contain additional heat exchange surfaces that reove uch of the reaining heat contained in the cobustion gases before these gases are exhausted to the exterior of the building. On one of these surfaces, cobustion gases are allowed to cool until water vapour condenses, which releases energy stored in the vapour. The increased efficiency is explained by the cobustion of propane to produce liquid water, releasing ore energy than a siilar process involving the production of water vapour (see above). A furnace that operates inefficiently exhausts cobustion gases that still contain a considerable aount of energy. Note that the incoplete cobustion of propane in part (c) released the least aount of energy. Coplete cobustion is increased in high-efficiency furnaces by directly providing cobustion air fro the outside. This (fresh cobustion air) feature reduces the quantity of war air being taken fro a hoe for cobustion. (It also reduces the possibility of a back-draft of carbon onoxide and carbon dioxide into the hoe fro the furnace chiney when a fireplace is burning and drawing significant quantities of air fro the hoe). 30. Prior to the advent of refrigerators, the Inuit stored and shared their eat in a counal fashion the eat had to be eaten before it spoiled. The advent of the individually owned refrigerator has destroyed this sense of counity sharing and dependence. Thus, while technological progress ay bring independence, it exacts a social cost. Extension 31. [Student responses will vary, depending on the alternative fuel chosen, the sources used, the perspectives exained, and the target audience. Possible choices of alternative sources of energy include tidal, solar/photovoltaic, bioass, and geotheral. Possible perspectives include environental, econoic, and social.] 32. (a) The efficiency of converting available cheical energy into electrical energy in theral electric power stations ranges fro 27.11% to 40.01%. Most fuels have a siilarly low efficiency for converting cheical energy into electrical energy around one-third (33%). Lignite is consistently the least efficient coal used in theral electric power stations. Light fuel oil (LFO) has a slightly lower percent efficiency than heavy fuel oil (HFO). The efficiency of wood used in theral electric power stations has steadily decreased over tie. (b) Non-cobustible energy sources include: solar energy photovoltaic/solar panels, theral electric (using giant arrays of irrors) and hydrolectric (using the water cycle) geotheral energy tidal energy wind energy, and nuclear energy 454 Unit 6 Solutions Manual Copyright 2007 Thoson Nelson