5.1 CHANGES IN MATTER AND ENERGY

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1 CHAPTER 5 THERMOCHEMISTRY Reflect on Your Learning (Page 296) 1. Absorbing energy: ice melting, water evaporating, photosynthesis. Releasing energy: water vapour condensing, respiration, combustion of gasoline. 2. Nuclear power plants, fossil fuel-burning power plants, hydroelectric power (largely from Niagara), solar power. 3. Both technologies drive turbines to generate electricity, but one uses falling water to drive the turbine and the other uses nuclear energy to create pressurized steam to drive the turbine. Try This Activity: Burning Food (Page 297) (a) Use q with student data. (b) The heat released equalled the difference in potential energy of the reactants and products. (c) Divide the heat by the mass of the nut. (d) The reactants would be the same but the nut might not be completely digested, so there would be products other than carbon dioxide and water. The energy might be stored rather than released to the surroundings as heat. The process would be more efficient in terms of production of energy in useful forms, rather than heat. (e) Much heat is lost to the surroundings, apart from the water in the can. Insulating the apparatus would improve results. 5.1 CHANGES IN MATTER AND ENERGY PRACTICE (Page 300) 1. (a) chemical (new products: water and carbon dioxide) (b) physical (no new products) (c) chemical (new products: water and carbon dioxide) (d) physical (no new products) (e) chemical (new products: hydrogen gas and zinc chloride) (f) physical (no new products) 2. System Surroundings (a) gas and oxygen air and metal parts (b) ice hand (c) gas and oxygen air and metal parts (d) wax stove (e) zinc and acid beaker and water solvent (f) ice body part 3. All of these systems may be regarded as open because energy and/or matter may escape from the system, often in the form of gases. However, (a) may be considered a closed system for the instant at which the air fuel mixture ignites. 4. The thimbleful has greater average thermal energy per molecule, but the pool has greater total thermal energy. 5. (a) exothermic (b) exothermic (c) endothermic Copyright 2003 Nelson Thermochemistry 159

2 Making Connections 6. (a) Answers could include biological examples such as: cellular respiration (chemical, exothermic, open); combustion of fuel examples such as burning gasoline (chemical, exothermic, open); change of state examples such as boiling water (physical, endothermic, open) or condensation (physical, exothermic, open); household processes such as the use of drain cleaner (physical, exothermic, open). (b) Exothermic reactions are more common. 7. (a) 1 calorie = 4.18 J; 1 Calorie = 1000 calories = 4.18 kj. (b) The reaction is controlled oxidation of food: a slow burn that releases energy. (c) Bomb calorimetry is used. PRACTICE (Page 302) 8. Aluminum has the lowest specific heat capacity, which implies that it requires the least amount of heat to change temperature, and will undergo the greatest temperature change. 9. q = 1.50 kg 4.18 kj/(kg C) ( ) C q = 506 kj q 10. m = (c T) = J/(4.18 kj/(kg C) 50 C) m = 383 g q 11. T = (m c) = J/(4000 g 3.5 J/(g C)) T = 18 C 12. (a) q = 100 kg 4.18 kj/(kg C) (45 10) C q = kj, or 15 MJ (b) money saved = 14.6 MJ /MJ 1500 = , or $ (a) q = 100 kg 4.18 kj/(kg C) (75 45) C q = kj, or 10 MJ (b) money saved = 10 MJ /MJ 1500 = , or $55 PRACTICE (Page 304) 14. When a change occurs in a system, the potential energy change ( H) of the system is numerically equal to the change in thermal kinetic energy (q) of the surroundings. 15. Chemical changes have generally greater enthalpy changes than physical changes; nuclear changes have much greater enthalpy changes than chemical changes. 160 Chapter 5 Copyright 2003 Nelson

3 Applying Inquiry Skills 16. Experimental Design A measured mass of the metal is added to a measured mass of dilute acid, and the temperature change in the solution is determined. The heat gained by the solution is calculated using q. The heat released per gram is calculated by dividing the heat by the mass of metal, and the result is compared to the three accepted values. SECTION 5.1 QUESTIONS (Page 305) 1. Changes of state from solid to liquid, liquid to gas, and solid to gas are endothermic; changes of state from gas to liquid, liquid to solid, and gas to solid are exothermic. 2. mass, temperature change, specific heat capacity 3. (a) chemical (new products; rearrangement of atoms to new molecules) (b) physical (no new product; same molecules) (c) nuclear (uranium decays to form new atoms) 4. System Surroundings (a) gasoline and oxygen engine block and air (b) water air and remaining water (c) uranium fuel concrete 5. (a) open (b) open (because there is no container keeping the water vapour in contact with the liquid water) (c) isolated (although open if one considers the waste heat produced) 6. Energies per mol for physical, chemical, and nuclear changes are on the order of 101, 103, and 1011 kj/mol, respectively. Making Connections 7. See the Nelson Chemistry 12 web site for possible useful sources of information. Bomb calorimeters are used to determine the energy content of foods, fuels, and even organisms in ecological food chains. 8. See the Nelson Chemistry 12 web site for possible useful sources of information. Cold packs typically contain ammonium salts which, when mixed with water, absorb energy. Some hot packs contain iron filings which slowly oxidize in air and produce heat. 5.2 MOLAR ENTHALPIES PRACTICE (Page 308) 1 mol 1. amount of water, n = g g n = 5.56 mol H = n H vap = 5.56 mol kj 1 l mo H = 227 kj Copyright 2003 Nelson Thermochemistry 161

4 2. molar mass of Freon, M = g/mol 1 mol n Freon = 500 g g n Freon = 4.14 mol H = n H vap = 4.14 mol kj/mol H = 145 kj 3. amount of water, n = mol g g n = mol H = n H vap = mol 6.03 kj/mol H = kj PRACTICE (Page 310) 4. = 150 g 4.18 J/g C ( ) C = 2320 J, or 2.32 kj molar mass of urea, M = 60.0 g/mol n urea = 10.0 g 1 mol 60 g n urea = mol n H solution = water urea H solution = q n kj mol H solution = 13.9 kj/mol Because the reaction is endothermic, H solution is kj/mol. 5. = 50.0 g 4.18 J/g C ( ) C = 794 J molar mass of gallium, M = g/mol n gallium = 10.0 g 1 mol/ g n gallium = mol n H solution = H solution = n q water gallium 794 J mol H solution = J/mol or 5.54 kj/mol Because the reaction is exothermic, H solution is 5.54 kj/mol. 162 Chapter 5 Copyright 2003 Nelson

5 Applying Inquiry Skills 5. Assume that 200 ml of solution is 200 g water. = 200 g 4.18 J/g C ( ) C = 5.94 kj amount of KOH = n = 5.2 g 1 mol/56.1 g n = mol n H reaction = H reaction = n 5.94 kj mol H reaction = 64 kj/mol Ba(NO 3 ) 2 Because the reaction is exothermic, H reaction is 64 kj/mol. 6. Answers will vary, but the student could use a polystyrene (Styrofoam) coffee cup calorimeter and thermometer to investigate temperature changes that occurred when the dextrose tablets were added to water. A mortar and pestle might be used to simulate the grinding process that occurs in chewing. Ambitious students might even investigate whether there was any effect of amylase (found in saliva) on the process. Making Connections 7. See the Nelson Chemistry 12 web site for possible useful sources of information. In general, the propane is used to vaporize and separate the components of an aqueous ammonia mixture. The ammonia gas then goes through cycles of condensation (outside the compartment, releasing heat to the air) and evaporation inside the compartment (absorbing heat from food). 5.3 REPRESENTING ENTHALPY CHANGES PRACTICE (Page 319) 1. (a) 2 C (s) + H 2(g) kj C 2 H 2(g) 2 C (s) + H 2(g) C 2 H 2(g) H = +228 kj H f = +228 kj/mol acetylene Endothermic Reaction products E p H reactants Reaction Progress Copyright 2003 Nelson Thermochemistry 165

6 (b) Al 2 O 3(s) kj 2 Al (s) + 3/2 O 2(g) Al 2 O 3(s) 2 Al (s) + 3/2 O 2(g) H = kj H decomp = kj/mol aluminum oxide Endothermic Reaction products E p H reactants Reaction Progress (c) C (s) + O 2(g) CO 2(g) kj C (s) + O 2(g) CO 2(g) H = kj H comb = kj/mol carbon Exothermic Reaction reactants E p H products Reaction Progress 2. (a) H comb = kj/mol H 2 (b) H comb = kj/mol NH 3 (c) H comb = kj/mol N 2 (d) H comb = kj/mol Fe 3. (a) H = 114 kj (b) H 2 SO 4(aq) + 2 NaOH (aq) Na 2 SO 4(aq) + 2 H 2 H = 114 kj (c) H neut = 114 kj/mol H 2 SO 4 (d) H neut = 57 kj/mol NaOH 4. (a) H 2(g) + 1/2 O 2(g) H 2 H = kj H 2 H 2(g) + 1/2 O 2(g) H = kj (b) Such equations have the same enthalpy change with a different sign. 5. (a) The reaction is exothermic because potential energy is converted to heat lost to the surroundings. (b) The reaction is endothermic because heat absorbed from the surroundings is converted to potential energy. 166 Chapter 5 Copyright 2003 Nelson

7 SECTION 5.3 QUESTIONS (Page 320) 1. (a) Since the reaction is exothermic, the PE diagram will resemble this. Reactants are octane and oxygen; the products are carbon dioxide and water. Exothermic Reaction reactants E p H products Reaction Progress (b) Since the reaction is endothermic, the PE diagram will resemble this. Reactants are boron and hydrogen; the product is diborane. Endothermic Reaction products E p H reactants Reaction Progress 2. (a) Mg (s) + O 2(g) + H 2(g) Mg(OH) (s) H = 925 kj (b) C 5 H 12(g) + 8 O 2(g) 5 CO 2(g) + 6 H 2 H = 2018 kj (c) NiO (s) Ni (s) + 1/2 O 2(g) H = 240 kj 3. (a) C 4 H 10(g) + 13/2 O 2(g) 4 CO 2(g) + 5 H MJ (b) C (graphite) + 2 kj C (diamond) (c) C 2 H O 2(g) 2 CO 2(g) + 3 H MJ Applying Inquiry Skills 4. Analysis (a) = g 4.18 J/(g C) ( ) C = 45.9(8) kj [Digit in parentheses will be lost in rounding.] Copyright 2003 Nelson Thermochemistry 167

8 q copper = 50.0 g J/(g C) ( ) C q copper = 1.06 kj n H reaction = q total = + q copper n H reaction = 47.0(4) kj [Digit in parentheses will be lost in rounding.] m eicosane = g m eicosane = 1.21 g M eicosane (C20 H 42 ) = 282 g/mol n eicosane = 1.21 g 1 mol 282 g n eicosane = mol H comb = q total n 47.0(4) kj mol H comb = kj/mol C 20 H 42 Because the reaction is exothermic, H reaction is 11.0 MJ/mol. (b) The reaction was exothermic, because heat was released to the surroundings and the temperature increased. (c) C 20 H 42(s) + 61/2 O 2(g) 20 CO 2(g) + 21 H MJ C 20 H 42(s) + 61/2 O 2(g) 20 CO 2(g) + 21 H 2 H = 11.0 MJ Evaluation (d) Percentage error = ( ) = 17% 5.4 HESS S LAW OF ADDITIVITY OF REACTION ENTHALPIES PRACTICE (Page 326) 1. 1 (1): 2 Al (s) O 2(g) Al 2 O 3(s) H = (1)( ) kj 1 (2): Fe 2 O 3(s) 2 Fe (s) O 2(g) H = ( 1)( 824.2) kj Fe 2 O 3(s) + 2Al (s) Al 2 O 3(s) + 2 Fe (s) H = kj (1): C (s) O 2(g) C H = ( 1 )( 221.0) kj (2): H 2 H 2(g) O 2(g) H = ( 1 )( 483.6) kj 2 H 2 + C (s) C + H 2(g) H = kj 168 Chapter 5 Copyright 2003 Nelson

9 (1): C C (s) O 2(g) H = ( 1/2)( 221.0) kj 1 2 (3): H 2(g) O 2(g) H 2 H = ( 1 )( 483.6) kj 2 1 (2): C (s) + O 2(g) CO 2(g) H = (1)( 393.5) kj C + H 2(g) + O 2(g) CO 2(g) + H 2 H = kj PRACTICE (Page 329) 4. 1 (1): C 2 H 2(g) O 2(g) 2 CO 2(g) + H 2 H = (1)( 1299) kj 2 (2): 2 H 2(g) + O 2(g) 2 H 2 H = (2)( 286) kj 1 (3): 2 CO 2(g) + 3 H 2 C 2 H 6(g) O 2(g) H = ( 1)( 1560) kj C 2 H 2(g) + 2 H 2(g) C 2 H 6(g) H = 311 kj Since the reaction is written for one mole of ethyne, the molar enthalpy of combustion of ethyne, H comb, is 311 kj/mol. 1 mol amount of ethyne, n = 200 g ethyne g n = 7.69 mol H = n H comb = 7.69 mol ( 311 kj/mol) H = kj (2): C C (s) O 2(g) H = ( 1 )( 221.0) kj 2 1 (3): CO 2(g) + 2 H 2 CH 4(g) + 2 O 2(g) H = ( 1)( 802.7) kj 1 (4): C (s) + O 2(g) CO 2(g) H = (1)( 393.5) kj 3 2 (1): 3H 2(g) O 2(g) 3H 2 H = ( 3 )( 483.6) kj 2 3 H 2(g) + C CH 4(g) + H 2 H = kj Since the reaction is written for one mole of carbon monoxide, the molar enthalpy of combustion, H comb,is kj/mol CO. amount of CO, n = 300 g 1 mol/ 28.0 g n = 10.7 mol H = n H comb = 10.7 mol ( kj/mol) H = kj Copyright 2003 Nelson Thermochemistry 169

10 SECTION 5.4 QUESTIONS (Page 330) 1. (a) The reference thermochemical equations are: (1) C 8 H 18(l) O 2 2(g) 8 CO 2(g) + 9 H 2 H 1 = 5470 kj (2) H 2(g) O 2(g) H 2 H 2 = kj (3) C (s) + O 2(g) CO 2(g) H 3 = kj (b) Applying Hess s Law: 1 (1): 8 CO 2(g) + 9 H 2 C 8 H 18(l) O 2 2(g) H = ( 1)( 5470) kj 9 (2): 9H 2(g) O 2(g) 9 H 2 H = (9)( 285.8) kj 8 (3): 8C (s) + 8 O 2(g) 8 CO 2(g) H = (8)( 393.5) kj 8 C (s) + 9 H 2(g) C 8 H 18(l) H = kj (1): HCl Na 2 O NaCl H 2 O H = ( 1 )(507) kj (2): NaNO NO NO Na 2 O H = ( 1 )( 427) kj (4): 1 2 N 2 O O H 2 O HNO 2 H = ( 1 )(34) kj (3): 1 2 NO NO N 2 O O 2 H = ( 1 )( 43) kj 2 HCl + NaNO 2 HNO 2 + NaCl H = 78.5 kj 3. The reference equations are: (1) C 2 H 5 OH (l) + 3 O 2(g) 2 CO 2(g) + 3 H 2 H 1 = 1367 kj (2) CH 3 COOH (l) + 2 O 2(g) 2 CO 2(g) + 2 H 2 H 2 = 875 kj Applying Hess s Law: 1 (1): C 2 H 5 OH (l) + 3 O 2(g) 2 CO 2(g) + 3 H 2 H = (1)( 1367) kj 1 (2): 2 CO 2(g) + 2 H 2 CH 3 COOH (l) + 2 O 2(g) H = ( 1)( 875) kj C 2 H 5 OH (l) + O 2(g) CH 3 COOH (l) + H 2 H = 492 kj Applying Inquiry Skills 4. (a) When reference equations (1) and (2) are added together, the result is reference equation (3). 1 (1): HBr (aq) + KOH (aq) H 2 + KBr (aq) H 1 1 (2): KOH (s) KOH (aq) H 2 (3) KOH (s) + HBr (aq) H 2 + KBr (aq) H 3 = H 1 + H 2 (b) In all three experiments, assume ml of solution is g water. 170 Chapter 5 Copyright 2003 Nelson

11 Experiment 1: = g 4.18 J/g C ( ) C = 2.09 kj n H 1 = = 2.09 kj amount of KOH, n = CV = 1.00 mol/l L n = mol q H 1 = n kj mol H 1 = 20.9 kj/mol KOH Because the reaction is exothermic and is written for one mole of KOH, H 1 is 20.9 kj. Experiment 2: = g 4.18 J/g C ( ) C = 3.43 kj n H 2 = = 3.43 kj molar mass of KOH, M = 56.1 g/mol 1 mol amount of KOH, n = 5.61 g g n = mol q H 2 = n kj mol H 2 = 34.3 kj/mol KOH Because the reaction is exothermic and is written for one mole of KOH, H 2 is 34.3 kj. Experiment 3: = g 4.18 J/g C ( ) C = 5.60 kj n H reaction = = 5.60 kj 1 mol amount of KOH, n = 5.61 g g n = mol q H 3 = n kj mol H 3 = 56.0 kj/mol KOH Because the reaction is exothermic and is written for one mole of KOH, H 3 is 56.0 kj/mol. Copyright 2003 Nelson Thermochemistry 171

12 ( H 1 + H 2 ) H 3 (c) Experimental percentage error H3 100% ( ) % = 1.4 % 5.5 STANDARD ENTHALPIES OF FORMATION PRACTICE (Page 332) 1. (a) 6 C (s) + 3 H 2(g) C 6 H 6(l) (b) K (s) Br 2(l) + 3/2 O 2(g) KBrO 3(s) (c) 6 C (s) + 6 H 2(g) + 3 O 2(g) C 6 H 12 O 6(s) (d) Mg (s) + H 2(g) + O 2(g) Mg(OH) 2(s) PRACTICE (Page 335) 2. (a) C 5 H 12(l) + 8 O 2(g) 5 CO 2(g) + 6 H 2 H f(c5 H 12(l) ) = kj/mol H f(o2(g) ) = 0 kj/mol H f(co2(g) ) = kj/mol H f(h2 ) = kj/mol H = 5 H f(co2(g) ) + 6 H f(h2 ) 1 H f(c5 H 12(l) ) 8 H f(o2(g) ) = 5 ( 393.5) + 6 ( 285.8) 1 ( 173.5) 8 (0) H = 3509 kj (b) Fe 2 O 3(s) + 3 C 2 Fe (s) + 3 CO 2(g) H f(fe2 O 3(s) ) = kj/mol H f(co(g) ) = kj/mol H f(fe(s) ) = 0 kj/mol H f(co2(g) ) = kj/mol H = 2 H f(fe(s) ) + 3 H f(co2(g) ) 1 H f(fe2 O 3(s) ) 3 H f(co(g) ) = 2 (0) + 3 ( 393.5) 1 ( 824.2) 3 ( 110.5) H = 24.8 kj 3. C 6 H 12(l) + 9 O 2(g) 6 CO 2(g) + 6 H 2 H comb(c6 H 12(l) ) = 3824 kj/mol H f(o2 (g) ) = 0 kj/mol H f(co2(g) ) = kj/mol H f(h2 ) = kj/mol 172 Chapter 5 Copyright 2003 Nelson

13 H comb(c6 H 12(l) ) = 6 H f(co2(g) ) + 6 H f(h2 ) ) 1 H f(c 6 H 12(l) ) 9 H f(o2 (g) ) 3824 = 6 ( 393.5) + 6 ( 285.8) 1 ( H f(c6 H 12(l) )) 9 (0) H f(c6 H 12(l) ) = 6 ( 393.5) + 6 ( 285.8) 9 (0) kj H f(c6 H 12(l) ) = 252 kj/mol The standard enthalpy of combustion of liquid cyclohexane is 252 kj/mol. 4. (a) H f(ch4(g) ) = 74.4 kj/mol H f(h2 ) = kj/mol H f(co(g) ) = kj/mol H f(h2(g) ) = 0 kj/mol H = 1 H f(co(g) ) + 3 H f(h2(g) ) 1 H f(ch4(g) ) 1 H f(h2 ) = 1 ( 110.5) + 3 (0) 1 ( 74.4) 1 ( 285.8) H = kj H for the reaction is kj. (b) H f(co(g) ) = kj/mol H f(h2 ) = kj/mol H f(co2(g) ) = kj/mol H f(h2(g) ) = 0 kj/mol H = 1 H f(co2(g) ) + 1 H f(h2(g) ) 1 H f(co(g) ) 1 H f(h2 ) = 1 ( 393.5) + 1 (0) 1 ( 110.5) 1 ( 285.8) H = 2.8 kj H for the reaction is 2.8 kj. (c) H f(n2(g) ) = 0 kj/mol H f(h2(g) ) = 0 kj/mol H f(nh3(g) ) = 45.9 kj/mol H = 2 H f(nh3(g) ) 1 H f(n2(g) ) 3 H f(h2(g) ) = 2 ( 45.9) 1 (0) 3 (0) H = 91.8 kj H for the reaction is 91.8 kj. 5. (a) H f(nh3(g) ) = 45.9 kj/mol H f(o2(g) ) = 0 kj/mol H f(no(g) ) = 90.2 kj/mol H f(h2 ) = kj/mol H = 4 H f(no(g) ) + 6 H f(h2 ) 4 H f(nh3(g) ) 5 H f(o2(g) ) = 4 (+90.2) + 6 ( 285.8) 4 ( 45.9) 5 (0) H = kj H for the reaction is kj. (b) H f(no(g) ) = 90.2 kj/mol H f(o2(g) ) = 0 kj/mol H f(no2(g) ) = 33.2 kj/mol Copyright 2003 Nelson Thermochemistry 173

14 H = 2 H f(no2(g) ) 2 H f(no(g) ) 1 H f(o2(g) ) = 2 (+33.2) 2 (+90.2) 1 (0) H = kj H for the reaction is kj. (c) H f(no2(g) ) = 33.2 kj/mol H f(h2 ) = kj/mol H f(hno3(l) ) = kj/mol H f(no(g) ) = 90.2 kj/mol H = 2 H f(hno3(l) ) + 1 H f(no(g) ) 3 H f(no2(g) ) 1 H f(h2 ) = 2 ( 174.1) + 1 (+90.2) 3 (+33.2) 1 ( 285.8) H = 71.8 kj H for the reaction is 71.8 kj. Making Connections 6. (a) raw materials + light energy + fertilizer (chemical energy) plant (chemical energy) prepared food (chemical energy) blood sugars (chemical energy) body tissue (chemical energy), movement (kinetic energy), or heat loss (thermal energy) (b) The advantage of using fertilizers to increase crop yield is that, although we input energy to make the fertilizers, the energy we gain back is now in a form that is usable by our bodies. The question of whether or not we should fertilize our crops goes far beyond a consideration of the amount of energy used to produce the fertilizer. It should include a weighing of the risks and benefits of both fertilizing (e.g., possibly polluting the ground water) and not fertilizing (e.g., producing less food). PRACTICE (Page 338) Making Connections 7. (a) H f(nh3(g) ) = 45.9 kj/mol H f(hno3(l) ) = kj/mol H f(nh4 NO 3(s) ) = kj/mol H = 1 H f(nh4 NO 3(s) ) 1 H f(nh3(g) ) 1 H f(hno3(l) ) = 1 ( 365.6) 1 ( 45.9) 1 ( 174.1) H = kj H for the reaction is kj. (b) Exothermic Reaction reactants E p H products Reaction Progress 174 Chapter 5 Copyright 2003 Nelson

15 (c) molar mass of NH 4 NO 3, M = 80.0 g/mol amount of NH 4 NO 3, n = mol g g n = mol q = H = n H r = mol kj q = kj As the reaction is exothermic, kj of heat will be lost to the surroundings during the reaction. 8. H f(coal(s) ) = kj/mol H f(o2 (g) ) = 0 kj/mol H f(co2(g) ) = kj/mol H f(h2 ) = kj/mol H = 104 H f(co2(g) ) + 16 H f(h2 ) 2 H f(coal(s) ) 111 H f(o2 (g) ) = 104 ( 393.5) + 16 ( 241.8) 2 ( 396.4) 111 (0) H = kj or 44.0 MJ H comb = nh MJ 2 mol coal H = 22.0 MJ/mol coal molar mass of C 52 H 16 O, M = 656 g/mol amount of C 52 H 16 O, n = g 1 mol 656 g n = 152 mol q = H = n H r = 152 mol 22.0 MJ/mol q = MJ Burning kg anthracite coal will produce MJ of thermal energy. 9. (a) CH 3 OH (l) O 2(g) 1 CO 2(g) + 2 H 2 H f(ch3 OH (l) ) = kj/mol H f(o2 (g) ) = 0 kj/mol H f(co2(g) ) = kj/mol H f(h2 ) = kj/mol H = 1 H f(co2(g) ) + 2 H f(h2 ) 1 H f(ch3 OH (l) ) 1.5 H f(o2 (g) ) = 1 ( 393.5) + 2 ( 285.8) 1 ( 239.1) 1.5 (0) H = 726 kj molar mass of CH 3 OH, M = 32.0 g/mol amount of CH 3 OH, n = mol g g n = 31.3 mol Copyright 2003 Nelson Thermochemistry 175

16 q = H = n H comb = 31.3 mol 726 kj/mol q = kj, or 22.7 MJ for one mole burned. (b) H 2(g) O 2(g) 1H 2 H f(h2(g) ) = 0 kj/mol H f(o2 (g) ) = 0 kj/mol H f(h2 ) = kj/mol H = 1 H f(h2 ) 1 H f(h2(g) ) 0.5 H f(o2 (g) ) = 1 ( 285.8) 1 (0) 0.5 (0) H = kj molar mass of H 2, M = 2.02 g/mol amount of H 2, n = mol g 2.02 g n = 495 mol q = H = n H comb = 495 mol kj/mol q = kj, or 142 MJ for one mole burned. (c) Both of these fuels produce less energy per mole than octane. (d) Cost of the fuels and availability are two economic factors. Consumer safety concerns are also an issue since hydrogen is viewed as a more explosive gas. Making Connections 10. (a) CH 4(g) + 2 O 2(g) 1 CO 2(g) + 2 H 2 H f(ch4(g) ) = 74.4 kj/mol H f(o2 (g) ) = 0 kj/mol H f(co2(g) ) = kj/mol H f(h2 ) = kj/mol H = 1 H f(co2(g) ) + 2 H f(h2 ) 1 H f(ch4(g) ) 2 H f(o2 (g) ) = 1 ( 393.5) + 2 ( 285.8) 1 ( 74.4) 2 (0) H = kj/mol methane Thus, H comb = kj/mol methane = g 4.18 J/(g C) (70 5) C = kj n H comb = ater comb q n = w H kj kj/mol n = 30 mol 176 Chapter 5 Copyright 2003 Nelson

17 m = n M = 30 mol 16.0 g/mol m = 480 g 480 g of methane will heat 100 kg of water from 5 C to 70 C. (b) Insulating water pipes and setting the thermostat at a lower temperature are ways to conserve energy. (c) Electricity and propane combustion are commonly used for heating water. SECTION 5.5 QUESTIONS (Page 339) 1. (a) 2 C (s) + H 2(g) C 2 H 2(g) (b) 4 C (s) + 9/2 H 2(g) + 3/2 N 2(g) + O 2(g) C 4 H 9 N 3 O 2(s) (c) K (s) + 1/2 I 2(s) KI (s) (d) Fe (s) + S (s) + 2 O 2(g) FeSO 4(s) 2. (a) MgCO 3(s) MgO (s) + CO 2(g) H = 1 H f( CO2(g) ) + 1 H f( MgO(s) ) 1 H f(mgco3(s) ) = 1 ( 393.5) + 1 ( 601.6) 1 ( ) H = kj (b) C 2 H 4(g) + 3 O 2(g) CO 2(g) + 2 H 2 H = 2 H f(co2(g) ) + 2 H f(h2 ) 1 H f(c2 H 4(g) ) 3 H f(o 2(g) ) = 2 ( 393.5) + 2 ( 285.8) 1 (+52.5) 3 (0) H = 1411 kj (c) C 12 H 22 O 11(s) + 12 O 2(g) 12 CO 2(g) + 11 H 2 H = 12 H f(co2(g) ) + 11 H f(h2 ) 1 H f(sucrose(s) ) 12 H f(o2(g) ) = 12 ( 393.5) + 11 ( 285.8) 1 ( ) 12 (0) H = 5640 kj 3. (a) C 8 H 18(g) + 3 H 2(g) CH 4(g) + 2 C 2 H 6(g) + C 3 H 8(g) (b) H = 1 H f(ch4(g) ) + 2 H f(c2 H 6(g) ) + 1 H f(c3 H 8(g) ) 1 H f(c8 H 18(g) ) 3 H f(h2(g) ) = 1 ( 74.4) + 2 ( 83.8) + 1 ( 104.7) 1 ( 250.1) 3 (0) H = 96.6 kj Applying Inquiry Skills 4. Prediction (a) C 3 H O 2(g) 3 CO 2(g) + 3 H 2 H = 3 H f(co2(g) ) + 3 H f(h 2 ) 1 H f(acetone(l) ) 4 H f(o2(g) ) = 3 ( 393.5) + 3 ( 285.8) 1 (248.1) 4 (0) H = kj or H c = 1.79 MJ/mol acetone Copyright 2003 Nelson Thermochemistry 177

18 Analysis (b) = g 4.18 J/g C ( ) C = 2.09 kj q aluminum = c T = 50.0 g 0.91 J/g C ( ) C q aluminum = 0.23 kj q total = + q aluminum + q aluminum = 2.32 kj n H c = q total m acetone = g M acetone = 58.0 g 1 mol n acetone = g g n acetone = mol H c = q total n 2.32 kj mol H c = 1.5 MJ/mol acetone Because the reaction is exothermic, the molar enthalpy of combustion of acetone, H c, is 1.5 MJ/mol. Evaluation (c) Percentage error = ( ) 100% 1.79 = 16% (d) The percentage error suggests that heat has been lost to the surroundings as part of experimental error. (e) If heat is lost to the surroundings, then the observed temperature change in the water and calorimeter, the calculated q values, and the experimental H all will be smaller than predicted. 5.6 THE ENERGY DEBATE PRACTICE (Page 344) 1. (a) hydroelectric power, nuclear power, burning fossil fuels (b) All produce power by driving turbines: nuclear and fossil fuels heat water to drive steam turbines, whereas hydroelectric power uses falling water. Hydroelectric and nuclear energy have higher capital costs. Hydroelectric is relatively environmentally benign, fossil fuel burning produces the greatest amount of pollution, and nuclear energy poses the greatest safety risk. Making Connections 2. (Sample answer) The CANDU system uses a heavy water moderator and ordinary uranium fuel, whereas some other systems use ordinary water as a moderator and enriched uranium fuel. 178 Chapter 5 Copyright 2003 Nelson

19 CHAPTER 5 SUMMARY MAKE A SUMMARY (Page 354) most energetic nuclear H+ least energetic chemical H endothermic physical types exothermic closed isolated open systems THERMO- CHEMISTRY energy changes heat flow H system = ±q surroundings changes measured using calorimetry H expressed as enthalpy changes determined by H = n H r q = mc T specific heat capacity Hess s Law H = Σ H known standard heats of formation H f Σn H fproducts n H freactants CHAPTER 5 SELF-QUIZ (Page 355) 1. False: Nuclear changes generally produce more energy than chemical changes. 2. False: In exothermic reactions, the reactants have more potential energy than the products. 3. True 4. False: In endothermic reactions, the heat term is written on the left side of the equation. 5. True 6. True 7. False: Burning of gasoline is an example of an exothermic physical change. 8. True 9. False: Specific heat capacity is the amount of heat required to change one gram through 1 C. 10. True 11. (c) 12. (b) 13. (e) 14. (c) 15. (c) 16. (c) 17. (e) 18. (c) Copyright 2003 Nelson Thermochemistry 185

20 Worked Answers: g 13. H vap (methanol) = k J = 37.8 kj/mol (e) m ol g 120 g kj = 246 kj (c) (2 mol 39.1 g/mol) CHAPTER 5 REVIEW (Page 356) 1. Physical Chemical Nuclear (a) change in state or change in change in arrangement of atoms change in arrangement of nuclei arrangement of atoms in molecules in molecules (b) about 10 kj/mol about 10 3 kj/mol about kj/mol (c) freezing water or melting butter burning gasoline or cooking food uranium decay or hydrogen fusion in Sun q 2. c m T J 938 g (35.0ºC 19.5ºC) c = 1.10 J/g ºC The specific heat capacity of the brick is 1.10 J/g ºC. 3. We assume that no heat is lost to the environment, negligible heat is lost to the calorimeter materials unless specific information is given about the container, and dilute aqueous solutions have density and specific heat capacity of water. 4. = 500 g 4.18 J/g ºC (80ºC 20ºC) = J q copper = 2000 g J/g ºC (80ºC 20ºC) q copper = J q total = + q copper q total = J, or 170 kj 170 kj of heat is required. 5. m water = d V = 1.00 g/ml g m water = g = g 4.18 J/g ºC (65 C 20 C) = kj 186 Chapter 5 Copyright 2003 Nelson

21 Worked Answers: g 13. H vap (methanol) = k J = 37.8 kj/mol (e) m ol g 120 g kj = 246 kj (c) (2 mol 39.1 g/mol) CHAPTER 5 REVIEW (Page 356) 1. Physical Chemical Nuclear (a) change in state or change in change in arrangement of atoms change in arrangement of nuclei arrangement of atoms in molecules in molecules (b) about 10 kj/mol about 10 3 kj/mol about kj/mol (c) freezing water or melting butter burning gasoline or cooking food uranium decay or hydrogen fusion in Sun q 2. c m T J 938 g (35.0ºC 19.5ºC) c = 1.10 J/g ºC The specific heat capacity of the brick is 1.10 J/g ºC. 3. We assume that no heat is lost to the environment, negligible heat is lost to the calorimeter materials unless specific information is given about the container, and dilute aqueous solutions have density and specific heat capacity of water. 4. = 500 g 4.18 J/g ºC (80ºC 20ºC) = J q copper = 2000 g J/g ºC (80ºC 20ºC) q copper = J q total = + q copper q total = J, or 170 kj 170 kj of heat is required. 5. m water = d V = 1.00 g/ml g m water = g = g 4.18 J/g ºC (65 C 20 C) = kj 186 Chapter 5 Copyright 2003 Nelson

22 UNIT 3 SELF-QUIZ (Page 412) 1. False: A physical change usually involves a larger enthalpy change than does a chemical change. 2. True 3. False: The potential energy of the products is smaller than the potential energy of the reactants in an exothermic change. 4. True 5. False: An endothermic reaction absorbs heat from the surroundings. 6. True 7. True 8. False: Three-quarters of a radioisotope will have changed after two half-lives. 9. False: In an endothermic reaction, only the potential energy of the chemical system increases. 10. True 11. (b) 12. (c) 13. (e) 14. (a) 15. (e) 16. (c) 17. (b) 18. (d) 19. (b) 20. (e) 21. (b) 22. (a) 23. (c) 24. (c) 25. (e) 26. (c) 27. (b) 28. (d) 29. (c) 30. (d) UNIT 3 REVIEW (Page 414) 1. = 1500 g 4.18 J/(g ºC) (75 20)ºC = 340 kj 2. M Cl2 = 70.9 g/mol 1 mol n Cl2 = 2250 g g n Cl2 = 31.7 mol H = n Cl2 H vap = 31.7 mol 20.7 kj/mol H = 657 kj 218 Unit 3 Copyright 2003 Nelson

23 3. (a) = g 4.18 J/(g ºC) 5.6ºC = 2341 J n NaOH = MV = mol/l L n NaOH = mol n H neut = H neut = n 2341 J mol H neut = 5573 J/mol, or 5.57 kj/mol Because the reaction is exothermic, H neut = 5.57 kj/mol. (b) The assumption is that the reaction went to completion and that all heat from the reaction was absorbed by the water in the calorimeter and not by the calorimeter or surroundings. 4. (a) 2 C (s) H 2(g) + 1/2 Cl 2(g) kj C 2 H 3 Cl (g) (b) 2 C (s) H 2(g) + 1/2 Cl 2(g) C 2 H 3 Cl (g) H = kj 5. (a) S (s) + O 2(g) SO 2(g) H = kj (b) Potential Energy Diagram of the Formation of Sulfur Dioxide enthalpy of S (g) + O 2(g) E p H = kj enthalpy of SO 2(g) Reaction Progress (c) M SO2 = g/mol = 9.63 g 1 mol/64.07 g n SO2 n SO2 = mol q = n SO2 H f = mol kj q = 44.6 kj Copyright 2003 Nelson Energy Changes and Rates of Reaction 219

24 6. 2 (3): 4 CH 3 NO 2(g) 4 C (s) + 6 H 2(g) + 4 O 2(g) + 2 N 2(g) H = ( 2) ( 226.2) kj 4 (1): 4 C (s) + 4 O 2(g) 4 CO 2(g) H = (4) ( 393.5) kj 3 (2): 6 H 2(g) + 3 O 2(g) 6 H 2 H = (3) ( 483.6) kj 4 CH 3 NO 2(g) + 3 O 2(g) 4 CO 2(g) + 2 N 2(g) + 6 H 2 H = kj 7. (a) C 5 H 12(l) + 8 O 2(g) 5 CO 2(g) + 6 H 2 (b) H = 5 H f(co2(g) ) + 6 H f(h2 ) 1 H f(c5 H 12(l) ) 8 H f(o2(g) ) = 5 ( 393.5) + 6 ( 285.8) 1 ( 146) 8 (0) H = kj (c) M pentane = 72.0 g/mol n pentane = 20 g 1 mol/ 72.0 g n pentane = mol q = n H comb = mol kj/mol q = 982 kj 982 kj would be released, when 20 g of pentane is burned. 8. Properties include colour, volume or pressure, and conductivity. 9. (a) As 1 mol of C is consumed, 1 mol of CO 2(g) is produced. At time 0, [CO 2(g) ] = 0 mol/l At time 40, [CO 2(g) ] = mol/l At time 100, [C ] = mol/l (b) As 1 mol of NO 2(g) is consumed, 1 mol of CO 2(g) is produced. In 80 s, [CO 2(g) ] = mol/l. Thus, [NO 2(g) ] = mol/l. [NO 2(g) ] 100 s = mol/l mol/l [NO 2(g) ] 100 s = mol/l 10. Newspapers, particularly as they curl while burning and because of air trapped in their pages, expose much more surface area to reaction with air. They therefore start burning more easily, and burn more quickly, than an equivalent quantity of wood. 11. (a) Graph of Decomposition of X 2 O 5(g) 1.4 Concentration (mol/l) [X 2 O 5(g) ] [O 2(g) ] Time (h) 220 Unit 3 Copyright 2003 Nelson

25 (b) As 1 mol of O 2(g) is produced, 4 mol of XO 2(g) are produced. Therefore, the four values for [XO 2(g) ] (in mol/l) are: 0.80, 1.30, 1.80, and (c) (i) [X 2 O t = 5(g) ] [X 2 O t = mol/(l h) In the first 12 h, the overall rate of consumption of X 2 O 5(g) is mol/(l h). (ii) + [ O2(g) ] 0.55 mol/l t = h + [ O2(g) ] t = mol/(l h) In the first 12 h, the overall rate of production of O 2(g) is mol/(l h). (iii) + [X O2(g) ] 2.20 mol/l t = h + [X O2(g) ] t = mol/(l h) In the first 12 h, the overall rate of production of XO 2(g) is mol/(l h). (d) When we use tangents to the curve, the rates of consumption of X 2 O 5(g) at 2.0 h and 7.0 h are 0.14 and mol/(l h), respectively. (e) The rate of consumption decreases as the concentration of reactant molecules able to collide and react decreases. 12. Aluminum powder has a much greater surface area, making it react much more quickly with oxygen in the air. 13. (a) The rate would double. (b) The rate would be halved. (c) Since both initial concentrations would be doubled, the rate would quadruple. 14. (a) This would have to be a multi-step mechanism because it involves six reactant molecules. (b) Rate of consumption of O 2(g) : [ O2(g) ] = mol/(l s) t [ O2(g) ] = mol/(l s) t Rate of production of CO 2(g) : + [ CO 2 ] = mol/(l s) t + [ CO 2 ] = mol/(l s) t 15. (a) When we compare Trials 2 and 3, we see that as [HI] is doubled, rate is multiplied by 2; therefore, rate depends on [HI] 1. (b) When we compare Trials 1 and 2, we see that as [O 2 ] is doubled, rate is multiplied by 2; therefore, rate depends on [O 2 ] 1. (c) The overall order is two. (d) r = k [HI] [O 2 ] r (e) k = [HI] [O 2 ] = 5(g) ] (mol/l s) mol/l mol/l ( ) mol/l 12.0 h k = 42 L/(mol s) (f) One molecule each of HI and O 2 are involved: two molecules. (g) Five reactant molecules are unlikely to collide in a single step at any appreciable rate. Copyright 2003 Nelson Energy Changes and Rates of Reaction 221

26 16. (a) 2 N + 2 H 2(g) N 2(g) + 2 H 2 (b) The reaction intermediates are N 2 O 2(g) and N 2. (c) r = k [NO] T 1 Number of Molecules T 2 > T 1 Kinetic Energy 18. Reaction (a) should be faster because it is homogeneous. In reaction (b), reactions can only occur at the surface of the lead metal. 19. (a) The activation energy barrier, E a, for white phosphorus is very low, whereas the E a for diamond is very high. (b) In the potential energy diagram for phosphorus, the activation energy barrier will be very small and the reaction exothermic with a large H; in the potential energy diagram for diamond, the activation energy barrier will be very large and the reaction exothermic with a very small H. Applying Inquiry Skills 20. Analysis (a) = 255 g 4.18 J/(g C) 28.8 C = 30.7 kj q copper = 305 g J/(g C) 28.8 C q copper = 3.38 kj + q copper = 34.1 kj n H reaction = q total m C3 H 6 O = 1.01 g M C3 H 6 O = 58.0 g n C3 H 6 O = 1.01 g 1 mol 58.0 g n C3 H 6 O = mol qt otal H comb n C3H6O 34.1 kj mol H comb = 1958 kj/mol propanal Because the reaction is exothermic, H comb = 1.96 MJ/mol. 222 Unit 3 Copyright 2003 Nelson

27 21. Analysis (a) Assume that ml of solution is g of water. = g 4.18 J/(g ºC) ( )C = 8.82 kj 1 mol n NaOH(s) = 3.40 g g n NaOH(s) = mol n HCl(aq) = MV = mol/l L n HCl(aq) = mol n H reaction = H reaction = n 8.82 kj mol H reaction = 104 kj/mol HCl Because the reaction is exothermic, H reaction is 104 kj/mol. Evaluation (b) (Answers will vary.) The Experimental Design is adequate if careful bomb calorimetry is used. 22. Analysis (a) The target equation is: 4 C (s) + 5 H 2(g) C 4 H 10(g) 1 (1): 4 CO 2(g) + 5 H 2 C 4 H 10(g) + 13/2 O 2(g) H = ( 1) ( ) kj 4 (2): 4C (s) + 4 O 2(g) 4 CO 2(g) H = (4) ( 393.5) kj 5 (3): 5H 2(g) O 2(g) 5 H 2 H = (5) ( 241.8) kj 4 C (s) + 5 H 2(g) C 4 H 10(g) H = kj Since this reaction is written for 1 mol of product, the molar enthalpy of formation of butane is kj/mol. 23. Experimental Design (a) The rate of reaction in terms of the rate of production of S will be measured by varying, one at a time, the initial concentrations of each of the reactants while controlling the other concentrations. Analysis (b) When we compare Trials 1 and 2, we see that as [A] is doubled, rate is multiplied by 2; therefore, rate depends on [A] 1. When we compare Trials 2 and 3, we see that as [B] is doubled, rate is multiplied by 2; therefore, rate depends on [B] 1. When we compare Trials 3 and 4, we see that as [C] is doubled, rate is multiplied by 1; therefore, rate depends on [C] 0. Overall, r = k [A] 1 [B] 1. Making Connections 24. (Answers will vary.) Nuclear reactors have higher capital cost, involve reactant fuels that require considerable processing, and have associated serious safety issues related to transportation and storage of radioactive isotopes. Fossil fuels are finite resources, which are nonetheless readily available. The capital cost of plants burning these fuels is relatively low and they can easily be converted to burn a range of fuels. However, fossil fuels are environmentally Copyright 2003 Nelson Energy Changes and Rates of Reaction 223

28 dirty energy sources that foul the air with, at best, greenhouse emissions and, at worst, acid-rain-producing sulfur dioxide, particularly when lower grades of coal are used. 25. (Answers will vary.) Carbon-14 and uranium-238 decay can be used to determine the ages of artifacts and ancient rocks. Medical research scientists are interested in the shelf life or rate of decomposition of antibiotics and the rate of retention of environmental toxins in the body after exposure. 26. (a) Increased concentration increases the collision frequency. (b) Decreased temperature decreases the fraction of molecules with enough kinetic energy to exceed the activation energy barrier. (c) Increased temperature increases the fraction of molecules with enough kinetic energy to exceed the activation energy barrier. (d) Reactions require activation energy in order to occur. (e) Increased surface area increases the collision frequency. (f) Addition of a catalyst lowers the activation energy barrier and increases the fraction of molecules with enough kinetic energy to exceed the activation energy barrier. (g) Homogeneous reactions allow much better mixing of reacting molecules, thus increasing the collision frequency. Extensions 27. (a) Efficient: C 8 H 18(l) O 2 2(g) 8 CO 2(g) + 9 H 2 Non-efficient: C 8 H 18(l) O 2 2(g) 4 CO 2(g) + 4 C + 9 H 2 (b) Efficient: H = 8 H f(co2(g) ) + 9 H f(h2 ) 1 H f(c8 H 18(l) ) 2 5 H 2 f(o2(g) ) = 8 ( 393.5) + 9 ( 285.8) 1 ( 250.1) 12.5 (0) H = 5470 kj Non-efficient: H = 4 H f(co2(g) ) + 4 H f(co(g) ) + 9 H f(h2 ) 1 H f(c8 H 18(l) ) 11.5 H f(o2(g) ) = 4 ( 393.5) + 4 ( 110.5) + 9 ( 285.8) 1 ( 250.1) 11.5 (0) H = 4338 kj (5470 kj 4338 kj) (c) % waste 100% 5470 kj % waste = 21% (d) = g 4.18 J/(g ºC) (95 ( 15))ºC = kj q block = g 0.50 J/(g ºC) (95 ( 15))ºC q block = kj q total = 1.7(9) 10 4 kj q total = n octane H comb qtota l n octane Hcom b = 1.7( 9) 10 4 kj 5470 kj n octane = 3.2(7) mol octane 224 Unit 3 Copyright 2003 Nelson

29 m octane = n octane M = 3.2(7) mol 114 g/mol m octane = 373 g, or g To raise the temperature of the engine to operating temperature, 373 g of octane must be burned. (e) You must assume that all of the energy produced by the burning of octane goes into the sum of temperature changes in the engine and motion of the vehicle. To drive 2 km, when warmed up, m octane = 2 km L/km 800 g/l m octane = 240 g To warm up the engine and drive 2 km, total m octane = total m octane = 613 g, or g 28. (Sample answer) In recent years, there have emerged worldwide infectious bacterial agents that are causing concern. These include: methicillin-resistant Staphylococcus aureus, multiple-drug resistant tuberculosis, and vancomycinresistant enterococci (VRE). There is extensive evidence that associates these resistances with the use, or abuse, of antibiotics in humans both in hospital and in the community. There is some suggestion that, with VRE, these resistances might originate from the several decades of use of animal feed additives. 29. Plant enzymes or meat marinades have been used for centuries to tenderize meats. For example, food might be wrapped in papaya leaves prior to cooking. Papain is an enzyme derived from these leaves, which catalyzes the breakdown of the muscle tissue, thus making the meat more tender. Figs and pineapples also contain protein-digesting enzymes which are produced as commercial meat tenderizers. These enzymes can only act on the surface of the meat, unless holes are poked in the surface, but the greater resultant fluid loss can be a problem in cooking. Most of these enzymes are most effective from 60 80ºC but are denatured and rendered ineffective at boiling temperature. 30. (a) H H H H H C O C H + O H 2 H C O H H H H (b) The acid is a catalyst. (c) CH 3 O CH 3 must be in the rate-determining step. (d) Any series of steps that uses up reactants, produces products, and consumes and regenerates the catalyst will be acceptable. For example, CH 3 O CH 3 + H + CH 3 OH + + CH 3 CH 3 OH + + H 2 O CH 3 OH + H 2 O + CH 3 + H 2 O + CH 3 OH + H + (e) Potential Energy Diagram of a Proposed Three-Step Mechanism for the Hydrolysis of Dimethyl Ether CH 3 O CH 3 + H + CH 3 OH + + CH 3 CH 3 OH + + H 2 O CH 3 OH + H 2 O + E p CH 3 O CH 3, H + CH 3 OH +, H 2 O CH 3 + H 2 O + CH 3 OH + H + CH 3, H 2 O + CH 3 OH, H + Reaction Progress Copyright 2003 Nelson Energy Changes and Rates of Reaction 225