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Solution Manual for Precalculus , 10th Edition

Solution Manual for Precalculus, 10th Edition offers the best solutions to textbook problems, helping you prepare for exams and assignments.

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    CHAPTER 2 1. a) q p = Δ H = 1 L × 0.9982 kg L –1 × 2447 kJ kg –1 = 2443 kJ b) Δ T = 2443 kJ/(60 kg × 4.184 kJ K –1 kg –1 ) = 9.7 K c) C 12 H 22 O 11 ( s ) + 12O 2 ( g ) → 12 CO 2 ( g ) + 11H 2 O ( l ) Δ H = 11(–285.83 kJ mol –1 ) + 12(–393.51 kJ mol –1 ) – (–2222.10 kJ mol –1 ) = –5644 kJ mol –1 M sucrose = 342.31 g mol –1 m sucrose = 2443 kJ × 342.31 g mol –1 /5644 kJ mol –1 = 148.2 g 2. a) 1 kg of carbon is (1.0 kg × 1000 g kg –1 )/(12.011 g mol –1 ) = 83.3 mol 83.3 mol CO 2 × 8.3145 J mol –1 K –1 × 298 K/(10 5 Pa) = 2.06 m 3 CO 2 2.06 m 3 CO 2 /(0.000390 m 3 CO 2 / m 3 air ) = 5292 m 3 air b) p = F/A = mg / A = 10 5 Pa. m = Ap / g = 1 m 2 × 10 5 Pa/(9.80662 m s –2 ) = 10197 kg M air = 0.8 × 0.0280 kg mol –1 + 0.2 × 0.0320 kg mol –1 = 0.0288 kg/(mol air) n air = 10197 kg/0.0288 kg/(mol air) = 354069 mol air n CO2 = 354069 mol air × 0.000390 mol CO 2 /(mol air) = 138.09 mol CO 2 m C = 0.012011 kg/mol C × 138.09 mol CO 2 = 1.659 kg C c) 1.659 kg C/(1 kg y –1 ) = 1.659 y. 3. a) w = mgh = 10 kg × 9.81 m s –2 × 10 m = 981 J b) k = F /( x – x 0 ) = 5.00 N/(0.105 m – 0.100 m) = 1.00 × 10 3 N m –1 w = ( k /2)( x – x 0 ) 2 = 1.00 × 10 3 N m –1 × (0.01 m) 2 /2 = 0.050 J Δ d / d surface × 100% = 1.2% c) w = – p ( V 2 – V 1 ) = –10 5 Pa × 0.002 m 3 = –200 J. d) w = –2 × 10 –4 J e) w = – nRT ln( V 2 / V 1 ) = – p 1 V 1 ln( V 2 / V 1 ) = –10 5 Pa × 10 –3 m 3 × ln 3 = –110 J. 4. a) w = f dx = 0 10 ∫ (20 x + x 2 ) dx = 0 10 ∫ 10 x 2 + 1 3 x 3 0 10 = (1000 + 333) × 10 − 21 J = 1.333 × 10 − 18 J b) w = 1.333 × 10 − 18 J × 6.022 × 10 23 mol –1 = 8.03 × 10 5 J mol –1 5. a) p = ρ gh = 1025 kg m –3 × 9.80662 m s –2 × 2500 m = 2.513 × 10 7 Pa = 251.3 bar b) Δ V / V = – κ Δ p = –(2.513 × 10 7 Pa) × 4.9458 × 10 − 10 Pa − 1 = –0.0124 So the density increases by a factor of 0.0124, or 1.24 % c) p 1 V 1 / T 1 = p 2 V 2 / T 2 , so V 2 = p 1 V 1 T 2 /( T 1 p 2 ) = 1 bar × 10 L × 277 K/(293 K × 252.3 bar)

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    Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak 2 = 0.0375 L. Density is reciprocally related to volume. 10/0.0375 = 266.9 – 1 = 265.9 = 26590% 6. n H 2 O = 100 g/(18.015 g mol –1 ) = 5.55 mol. a) q = 5.55 mol × 75.4 J mol –1 K –1 × 100 K = 4.19 × 10 4 J b) q = n H 2 O Δ fus H = 5.55 mol × 6007 J mol –1 = –3.33 × 10 4 J c) q = n H 2 O Δ vap H = 5.55 mol × 40667 J mol –1 = 2.26 × 10 5 J 7. a) V 1 = 1 mol × 8.3145 J mol –1 K –1 × 300 K /(10 5 Pa) = 0.0249 m 3 V 2 = 1 mol × 8.3145 J mol –1 K –1 × 600 K /(10 5 Pa) = 0.0499 m 3 w p = – p ( V 2 – V 1 ) = –10 5 Pa × (0.0499 – 0.0249)m 3 = –2494 J b) Δ U = nC V ,m Δ T = 1 mol × 20.8 J mol –1 K –1 × 300 K = 6.24 kJ Δ H = Δ U + Δ ( pV ) = 6.24 kJ + 2.49 kJ = 8.73 kJ c) q p = Δ H = 8.73 kJ 8. a) The volume, V = (4/3) π r 3 = 1.333 × 3.142 × (5 × 10 –7 m) 3 = 5.236 × 10–19 m 3 m = V ρ = 5.236 × 10–19 m 3 × 1000 kg m –3 = 5.236 × 10–16 kg. b) M = mN A = 5.236 × 10–16 kg × 6.022 × 1023 mol–1 = 3.153 × 108 kg mol–1 (kD) c) N water = M/M water = 3.153 × 108 kg mol–1/0.018 kg mol–1 = 1.752 × 10 10 d) 6378.1 × 103 m/(1 × 10 –6 m) = 6.3781 × 1012 e) A = 4 π r 2 = 4 × 3.142 × (5 × 10 –7 m) 2 = 3.142 × 10–12 m 2 f) 3.142 × 10–12 m 2 /(0.50 × 10–18 m 2 ) = 6.283 × 106 9. a) Δ U = Δ H = 0. w T = p ( V 2 – V 1 ) = p × nRT (1/ p 2 – 1/ p 1 ) = 10 5 Pa × (1 mol)(8.3145 J K –1 mol –1 )(298 K)[1/(105 Pa) – 1/(106 Pa)] = 2230 J q T = – w T b) q = 1 mol × (5/2) × 8.3145 J mol–1 K–1 × 75 K = 1559 J w = 0 Δ U = q Δ H = Δ U + nR Δ T = 1559 J + 1 mol × 8.3145 J mol–1 K–1 × 75 K = 2183 J c) p = 1 bar × 373 K /(298 K) = 1.252 bar d) lower, because the system will do work against the surroundings. 10. a) q = 0, adiabatic. w < 0, expansion. Δ U = w < 0. Δ H < 0. b) Δ U = Δ H = 0 since Δ T = 0

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    Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak 3 w < 0, expansion q = – w > 0 c) q = 0, adiabatic. w = 0, expansion against zero pressure Δ U = w + q = 0. Δ H = Δ U + R Δ T = 0 d) q > 0, heat of vaporization. w < 0, expansion against pressure Δ H = q p > 0. Δ U = Δ H – Δ ( pV ) > 0; Δ H >> Δ ( pV ) ≅ n g RT e) q < 0, exothermic reaction. w = 0, closed bomb, Δ V = 0 Δ U = q V < 0. Δ H = q V + Δ ( pV ) < 0; Δ ( pV ) = –( n H 2 + n O 2 ) RT < 0 11. a) Δ H = q p = 40.667 kJ. Δ U = Δ H – Δ ( pV ) = Δ H – Δ ( n g RT ) = 40667 J – (1 mol × 8.3145 J mol –1 K –1 × 373.15 K) = 37.564 kJ w p = p Δ V = Δ ( n g RT ) = 1 mol × 8.3145 J mol –1 K –1 × 373.15 K = –3.103 kJ b) w = w 1 + w 2 = 0, since p ex = 0 in step 1 and Δ V = 0 in step 2. q = Δ U = 37.564 kJ, path independent. Δ H = 40.667 kJ, path independent. 12. a) q = nC m ( T 2 − T 1 ) = (1 mol)(38.0 J K − 1 mol − 1 )(20 K) = 760 J b) q = (1 mol)(6007 J mol − 1 ) = 6007 J c) Per degree temperature drop, 10 mol H 2 O ( l ) loses (10 mol)(75.4 J mol − 1 K − 1 ) = 754 J K − 1 760 J + 6007 J + (1 mol)(75.4 J K − 1 mol − 1 )( T f − 0) = (10 mol)(75.4 J K − 1 mol − 1 )(20 − T f ) T f = (10)(75.4)(20) − 760 − 6007 (11)(75.4) = 10.0 ° C The final state is liquid. 13. a) Δ T = q P nC p ,m = − 400 J (36.5 J K − 1 mol − 1 × 1 mol) = –11.0 K T f = 125.0 − 11.0 = 114.0 ° C There is no phase change, so Δ V = nR Δ T / p = 1 mol(8.314 J K − 1 mol − 1 )( − 11.0 K) 10 5 Pa = − 9.1 × 10 − 4 m 3

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    Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak 4 b) Δ T = − 400 J (1 mol × 75.95 J mol − 1 K − 1 ) = − 5.27 °C T f = 100 − 5.27 = 94.73 ° C V f ( l ) = (18 g)/(0.96 g mL − 1 ) = 18.75 mL . No phase change; ∆ V is negligible for the liquid water. c) Δ n g = − 400 J ( ) 40,657 J mol − 1 ( ) = − 9.84 × 10 − 3 mol ∆ n g mol water vapor will condense to liquid No temperature change Δ V = Δ nRT / p = ( − 9.84 × 10 − 3 mol)(8.314 J mol -1 K -1 )(373 K) / (10 5 Pa) Δ V = − 3.05 × 10 − 4 m 3 d) Part (a), because it has the greatest Δ V ; w > 0. 14. V = 5.8 – 2 × exp[100(0.002)] =3.36 L At 1.005 bar, V = 5.8 – 2 × exp[100(–0.005)] =4.59 L Δ n = 10 5 Pa × (4.59 – 3.36) × 10 –3 m 3 /(8.3145 J mol –1 K –1 × 310 K) = 0.0477 mol 15. a) q = 0, thermally insulated w = 0, assuming no volume change because decrease in volume of hot copper equals increase in cool copper. Δ U = 0, energy is conserved. Δ H = 0 b) q = 0, thermally insulated w = 0, mechanical work done on the system Δ U = w > 0 Δ H = 0 because Δ pV = 0, with negligible expansion of the system. c) q = 0 w = 0 Δ U = 0 Δ H = 0 (all answers assume ideal gas behavior) 16. a) First Law of Thermodynamics – no restrictions b) Constant pressure process with no non- pV work. c) Enthalpy is a linear function of temperature (heat capacity is constant). i.e. Δ H / Δ T = dH / dT . d) Reactions in ideal gases at constant temperature, where Δ ( pV ) = RT Δ n e) van der Waals gas. f) Constant p ex .

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    Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak 5 17. c = C p ,m / M = 75.4 J mol –1 K –1 /(0.018 kg mol –1 ) = 4189 J/K. So 4 kJ/hr corresponds to almost 1° K per hour. It’s probably best to limit moon walks to two hours, since 2°C (3.6 F) above normal is a substantial fever. 18. n = pV RT = 10 5 Pa ( ) 5 × 10 − 4 m 3 ( ) 8.314 J K − 1 mol − 1 ( ) 293 K ( ) = 0.021 mol Per breath, Δ H = nC p ,m ( T 2 – T 1 ) = 0.021 mol × 3.5 × 8.3145 J mol –1 K –1 × 17.15 K = 10.2 J. Per day, 10.2 J × 12 min –1 × 60 min hr –1 × 24 hr d –1 = 177 kJ d –1 , representing about 1.5% of food energy. Metabolic heat = 80 kg × 4 kJ h –1 kg –1 × 24 h = 7.68 kJ d –1 , or 0.06% of food energy. At outside temperatures of –40°C, 4.5 × more energy will be lost through breathing; 7% of food energy begins to be significant. 19. a) w=p Δ V = 10 5 Pa × 5 × 10 –4 m 3 = 50 J/breath. In 24 hr, w = 50 J × 15000 d –1 = 750 kJ b) w = mgh = 750 kJ so m = w / gh = 750 kJ/(9.8 m s –2 × 100 m) = 765 kJ. 20. By convention, lower case symbols are specific (per kilogram) rather than molar quantities m water c p ,m,water T f − 55 ( ) K + m ice c p ,m,ice × 10 K + m ice ( ) Δ fus h + m ice c p ,m,water T f – 0 ( ) K = 0 0.100 kg × 4.184 kJ K − 1 kg − 1 × T f − 55 ( ) K + 0.010 kg × 2.11 kJ K − 1 kg − 1 10 K ( ) + 0.010 kg × 333.4 kJ kg − 1 + 0.010 kg × 4.184 kJ K − 1 kg − 1 × T f K = 0 0.4184 T f − 23.01 + 0.211 + 3.334 + 0.04184 T f = 0 T f = 19.47 0.460 = 42.3 ° C 21. Δ H 298 ° = 6 Δ f H ° CO 2 , g ( ) + 6 Δ f H ° H 2 O,l ( ) − Δ f H ° C 6 H 12 O 6 , s ( ) − 6 Δ f H ° O 2 , g ( ) = [6 − 393.509 ( ) + 6 − 285.830 ( ) − − 1274.4 ( ) − 0] kJ mol − 1 = − 2801.6 kJ mol − 1 22. a) q p = Δ H = 2 Δ f H ° ethanol, l ( ) + 2 Δ f H ° CO 2 , g ( ) − Δ f H ° glucose, s ( ) = 2 − 276.98 ( ) + 2 − 393.509 ( ) − − 1274.4 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 = − 66.6kJ b) This represents 2.4% of the heat available from the complete combustion. 23. Δ r H ° = Δ f H ° H 2 O, l ( ) + 1 2 Δ f H ° O 2 , g ( ) − Δ f H ° H 2 O 2 , aq ( )

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    Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak 6 Δ r H ° = ( − 285.830) + 0 − ( − 191.17) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 = − 94.66 kJ mol − 1 A temperature rise of 0.02 ˚C requires (0.02 ˚C)(4.184 kJ kg –1 )(1 kg L –1 ) = 0.0837 kJ L –1 Minimum detectableconcentration = 0.0837 kJ L − 1 94.66kJ mol − 1 = 8.84 × 10 − 4 M 24. a) Δ H ° 298 = − 74.81 ( ) + 0 − − 238.57 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 = 163.76 kJ mol − 1 b) Δ U ° = Δ H ° − Δ n gases RT = 163.76 kJ mol − 1 − 3 2 8.3145 J mol − 1 K − 1 1000 ( ) 298 K ( ) = 160.04 kJ mol − 1 c) Δ H ° 773 = Δ H ° 298 + Δ C p dT 298 773 ∫ where Δ C p = C p ,m CH 4 ( ) + 1 2 C p ,m O 2 ( ) − C p ,m CH 3 OH ( ) 25. Δ r H ° 298 = Δ f H ° (urea, s ) + 3 Δ f H ° (CO 2 , g ) + 3 Δ f H ° (H 2 O, l ) − 2 Δ f H ° (glycine, s ) Δ r H ° 298 = − 333.17 ( ) + 3 − 393.509 ( ) + 3 − 285.83 ( ) − 2 − 537.2 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 1 mol kJ 8 . 1296 − − = 26 a) Δ r H ° 298 = 2 − 276.98 ( ) + 2 − 393.509 ( ) − − 1274.4 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 = − 66.6 kJ mol − 1 b) Δ r H ° 298 = 2 − 484.1 ( ) + 2 − 393.509 ( ) + 2 − 285.830 ( ) − − 1274.4 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 1 mol kJ 5 . 1052 − − = c) Δ H 298 ° = − 2801.6 kJ mol − 1 (see Prob. 21) 27. a) From problem 26 a) Δ r H ° 353 = Δ r H ° 298 + Δ T Δ r C p = –66.6 kJ mol –1 + (353 – 298)K(2 × 111.5+2 × 37.1–219.2)J mol –1 K = –62.3 kJ mol –1 b) From problem 26 b) Δ r H ° 353 = –1052.5 kJ mol –1 + (353 – 298)K(2 × 123.5+2 × 75.4+2 × 37.1–219.2–2 × 29.4)J mol –1 K = –1041.8 kJ mol –1

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    Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak 7 28. a) Δ r H ° 298 = Δ f H ° sucrose, s ( ) + 12 Δ f H ° O 2 , g ( ) − 12 Δ f H ° CO 2 , g ( ) − 11 Δ f H ° H 2 O, l ( ) = − 2222.1 ( ) + 0 − 12 − 393.509 ( ) − 11 − 285.830 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 = 5644.1 kJ mol − 1 b) 20 kgsucrose ha − 1 hr − 1 = 5644.1 kJ mol − 1 ( ) 0.3423 kg mol − 1 ( ) − 1 20 kg ( ) = 329,800 kJ h − 1 = 91.6 kW ha − 1 c) Stored energy = 91.6 kW hectare − 1 ( ) 10 − 4 hectare m − 2 ( ) 1 kW m − 2 [ ] 100 % 92 . 0 = 29. a) H 2 g ( ) + 1 2 O 2 g ( ) → H 2 O l ( ) Δ r H ° 298 = Δ f H ° H 2 O, l ( ) = − 285.830 kJ mol − 1 ( ) 2.016 g H 2 mol − 1 ( ) = − 141.8 kJ g H 2 ( ) − 1 b) C 8 H 18 g ( ) + 25 2 O 2 g ( ) → 8CO 2 g ( ) + 9H 2 O l ( ) Δ r H 298 ° = − − 208.45 ( ) − 0 + 8 − 393.509 ( ) + 9 − 285.830 ( ) ⎡ ⎣ ⎤ ⎦ 114 gC 8 H 18 mol ( ) = − 48.35 kJ gC 8 H 18 ( ) − 1 c) Δ r H 298 ° H 2 ( ) Δ r H 298 ° C 8 H 18 ( ) = − 141.8 − 48.35 = 2.9 Thus H 2 is nearly 3 times more energetic as a fuel than octane on a weight basis. 30. a) Δ r H ° = Δ f H ° Ala ( ) + Δ f H ° CO 2 ( ) − Δ f H ° Asp ( ) = − 562.7 ( ) + − 393.509 ( ) − − 973.37 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 = 17.2kJ mol − 1 , absorbed Heat capacities needed Δ r H 323 ° = Δ r H 298 ° + Δ r C p ,m 50 − 25 ( ) where Δ r C p ,m = C p ,m Ala ( ) + C p ,m CO 2 ( ) − C p ,m Asp ( )

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    Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak 8 31. To measure Δ r H 298 ° the reactants could be put into a calorimeter at 25 ˚C and a small amount of an enzyme catalyst added. The heat evolved per mole at constant pressure is Δ r H 298 ° To estimate Δ r U 298 ° the volume change of the reaction must be determined. The change in volume can be measured in a dilatometer, or partial molal volumes can be used to calculate the change in volume. For practical purposes the volume change is negligible, since no gases are involved, and Δ r U 298 ° = Δ r H 298 ° . 32. a) 22 kWhd − 1 ( ) 3600 kJ (kWh) − 1 ( ) = 7.92 × 10 4 kJ d − 1 b) 1 kW m − 2 ( ) 0.1 ( ) 5 h ( ) = 0.5 kWh m − 2 A = 22 kWh ( ) 0.5 kWh m − 2 ( ) = 44 m 2 33. a) Δ r H 298 ° = 2 Δ f H ° H 2 O, g ( ) + Δ f H ° O 2 , g ( ) − Δ f H ° H 2 O 2 , g ( ) = 2 − 241.818 ( ) + 0 − 2 − 133.18 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 = − 217.28 kJ mol − 1 b) ( ) ( ) ( ) 1 mol kJ 5 . 140 3 . 498 2 1 28 . 217 2 1 O O D − = + − = − c) Δ r H ° 298 = 2 − 285.830 ( ) + 0 − 2 − 191.17 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 = − 189.32 kJ mol − 1 d) Heat evolved = 1 2 189.32 kJ (mol H 2 O 2 ) − 1 ( ) 0.01 molkg − 1 ( ) = 0.947 kJ kg − 1 Temperature rise = 0.947 kJ kg − 1 ( ) 4.18 kJ K − 1 kg − 1 ( ) = 0.226 K Final temperature 25.226 ° C 34. a) 8 C graphite ( ) + 9H 2 g ( ) → C 8 H 18 g ( ) Δ f H ° = 8 716.7 ( ) + 9 436.0 ( ) − 7 359 ( ) − 18 411 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 = − 253 kJ mol − 1 Published value = –208.45 kJ mol –1

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    Chapter 2 solutions manual to accompany 5 th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak 9 Chief discrepancy lies in using average bond dissociation energies for D(C—C) and D(C—H) b) 10 C (graphite) + 4 H 2 ( g ) → Δ f H ° = 10 Δ f H ° C ( ) + 4D H − H ( ) − 6D C − C ( ) − 5D C = C ( ) − 8D C − H ( ) = 10 716.7 ( ) + 4 436.0 ( ) − 6 359 ( ) − 5 611 ( ) − 8 411 ( ) ⎡ ⎣ ⎤ ⎦ kJ mol − 1 = 414 kJ mol − 1 Published value 1 mol kJ 96 . 150 − Chief discrepancy arises from neglect of resonance energy. c) C graphite ( ) + H 2 g ( ) + 1 2 O 2 g ( ) → H 2 C = O g ( ) Δ f H ° = Δ f H ° C ( ) + D H − H ( ) + 1 2 D O 2 ( ) − 2D C − H ( ) − D C = O ( ) = 716.7 + 436.0 + 498.3 / 2 − 2 411 ( ) − 709 = − 129 kJ mol − 1 Published value 1 mol kJ 90 . 115 − − Chief discrepancy arises from use of average bond dissociation energies. d) C (graphite) + H 2 ( g ) + O 2 ( g ) → formic acid Δ H f ° = Δ H f C ( ) + D H − H ( ) + D O 2 ( ) − D C − H ( ) − D C = O ( ) − D C − O ( ) − D O − H ( ) = 716.7 + 436.0 + 498.3 − 411 − 709 − 361 − 452 = − 282 kJ mol − 1 Published value − 424.76 + 46.15 = − 378.61 kJ mol − 1 Chief discrepancy comes from neglect of hydrogen bonding. 35. The steric repulsion can be estimated by the enthalpy change for the process cis -2-butene → trans -2-butene Δ r H ° 298 = − 11.1 kJ mol − 1 − − 7.0 kJ mol − 1 ( ) = − 4.1 kJ mol − 1 by which we see that the trans form has the lower enthalpy by 4.1 kJ/mol.

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    3-14 CHAPTER 3 1 . a) path 1 : (1 mole ideal gas, 3 bar, 450K)  (1 mol, 3 bar, 250 K)  (1 mol, 2 bar, 250 K) path 2 : (1 mole ideal gas, 3 bar, 450K)  (1 mol, 2 bar,450 K)  (1 mol, 2 bar, 250 K) path 1: the first step is isobaric, the second is isothermal; both reversible path 2: the first step is isothermal, the second is isobaric; both reversible work(path 1) =- nR (250-450K) + nR (250K)ln(2bar 3 bar) = -(1mol × 8.314J mol − 1 K − 1 )( − 200K) + (1mol × 8.314J mol − 1 K − 1 )(250K)ln(2bar 3 bar) = 820.0408 J q (path 1) = nC p (250 − 450K) − nR (250K)ln(2bar 3bar ) = (1mol × 2.5*8.314J mol − 1 K − 1 )( − 200K) − (1mol × 8.314J mol − 1 K − 1 )(250)ln(2 3 ) = − 3314.2408 J work(path 2) = nR (450K)ln(2bar 3 bar)- nR (250-450K) = (1mol × 8.314J mol − 1 K − 1 )(450K)ln(2bar 3bar) − (1mol × 8.314J mol − 1 K − 1 )( − 200) = 145.8334 J q (path 2) = − nR (450K)ln(2bar 3 bar) + nC p (250 − 450K) = − (1mol × 8.314J mol − 1 K − 1 )(450K)ln(2bar 3bar ) + (1mol × 2.5 × 8.314J mol − 1 K − 1 )( − 200K) = − 2640.0334 J (excess digits are shown since values will be rounded in part b) b) Δ U (path 1) = q path 1 + w path 1 = 820.0408 − 3314.2408 = − 2494.2 J (first law) Δ U (path 2) = q path 2 + w path 2 = 145.8334 − 2640.0334 = − 2494.2 J Δ S (path 1) = nC p ln(250K 450 K) − nR ln(2 bar 3 bar) = (1mol × 2.5 × 8.314 J mol -1 K -1 )*ln(250 450 ) − (1mol × 8.314 J mol -1 K -1 )*ln(2bar 3 bar ) = − 8.85 J K -1 Δ S (path 2) = − nR ln(2bar 3 bar) + nC p ln(250K 450K ) = − (1mol × 8.314 J mol -1 K -1 )*ln(2bar 3 bar) + (1mol × 2.5 × 8.314 J mol -1 K -1 )*ln(250 K 450 K) = − 8.85 J K -1

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    Chapter 3 solutions manual to accompany 5’th Edition of Physical Chemistry , Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak 2 2 2. For a heat engine to have a good efficiency there must be a significant difference in the hot and cold baths that it utilizes. Temperatures are very uniform on the surface of the earth so that organisms have not been exposed to simultaneous temperature differences in their environment thus there can be no evolutionary pressure for a heat engine to come about. Of course, extremely high temperatures would be needed for the hot bath in order to achieve a good efficiency and it is unlikely that any stable molecules of life could evolve or be stable at such high temperatures. 3 . a) kJ 3 . 133 (0.75) kJ) (100 ; 75 . 0 Eff 1 1 = = = − = q q w (Eq. 3.3) kJ 3 . 33 kJ) 3 . 133 ( kJ) 100 ( ; 1 3 3 1 − = − = − − = − = + q w q w q q 1 q (positive) is heat absorbed by the system at the hot temperature. 3 q (negative) is heat discharged by the system at the low temperature. w (negative) is the work done by the system. b) 1 1 3 1 25 . 0 kJ 100 q q q q w + − = − − = + = kJ 3 . 33 and kJ 3 . 133 Thus 3 1 + = − = q q Heat 3 q (positive) is absorbed by the system from a cold reservoir and heat 1 q (negative) is discharged to the hot reservoir when work w (positive) is done on the system. The engine is acting as a refrigerator. c) kJ 125 (0.80) kJ) (100 Eff 1 = = ′ − = ′ w q kJ 25 125 100 1 3 − = − = ′ − ′ − = ′ q w q d) Net heat absorbed at kJ 3 . 8 25 3 . 33 ´ 3 3 cold = − = + = q q T Net heat discharged at kJ 3 . 8 125 3 . 133 ´ 1 1 hot − = + − = + = q q T No net work done. In this hypothetical reversible cycle heat has been transferred from a cold reservoir to a hot reservoir with no input of work. This cannot happen. e) Assume an efficiency of 0.6. Then 6 . 0 1 = ′ ′ ′ ′ − q w For kJ 67 kJ; 167 (0.60) kJ) 100 ( kJ, 100 3 1 + = ′ ′ − = − = ′ ′ + = ′ ′ q q w Engine (a) could then be used to drive this one in reverse, as a refrigerator. As a result hot 1 1 cold 3 3 at discharged heat kJ, 34 167 133 at absorbed heat kJ, 34 67 33 T q q T q q − = − = ′ ′ + = + − = ′ ′ + No net work done. Again, an impossible result occurs. 4 . The organisms in the culture grow by using chemical energy from the nutrient medium. The gain in entropy from breakdown of complex foodstuffs into much simpler, higher entropy molecules, more than compensates for the loss in entropy associated with the “growth” of the organism. Metabolic processes also release heat into the surroundings, further raising the entropy of the surroundings.

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    Chapter 3 solutions manual to accompany 5’th Edition of Physical Chemistry , Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak 3 3 5 . a) Δ S mix = Δ S A + Δ S B = − n A R ln( p A 1bar) − n B R ln( p B 1bar) from Eq. (3.21) Now n A + n B = 1;therefore X A = n A ( n A + n B ) = P A (1bar) and X B = n B ( n A + n B ) = p B (1bar) Δ S m,mix = Δ S mix ( n A + n B ) = − X A R ln X A − X B R ln X B b) Δ S m,mix > 0 , because X A < 1 and ln X A < 0 , etc A positive value for Δ S m, mix means that the process should occur spontaneously in an isolated system, which is what we expect for the mixing process. c) At constant temperature, Δ G m,mix = Δ H m,mix − T Δ S m,mix (Eq. 3.32) Δ H m, mix can be positive, negative or zero. For ideal gases or for the formation of an ideal liquid solution, Δ H m, mix = 0 . In this case Δ G m, mix = − T Δ S m, mix < 0 , which is what we expect for the spontaneous mixing process. 6 . a) The amount of heat released when O(g) H 2 condenses raises the temperature to 100 ° C . b) 0 on) condensati ( ) ( = Δ + Δ = Δ = H T q C H q p p (0.20 mol)(36.5 J mol -1 K -1 )(5 K) + ( n mol)( − 40.66 kJ mol − 1 ) = 0 n = no. of mols of H 2 O(g) condensed = 8.98 × 10 − 4 mol where the heat capacity of steam at 100˚C has been approximated as that at 99.6˚C and 1 bar (table 2.2) c) Δ S = nC p (g)ln T 2 T 1 + Δ S (condensation) = (0.20 mol)(.0365 kJ K − 1 mol − 1 )ln 373 368 + (8.977 × 10 − 4 mol)( − 40.66kJ mol − 1 ) 373K Δ S = 6.60 × 10 − 7 kJ K − 1 7 . (a) This is a statement of the Second Law. If two reversible heat engines have different efficiencies then they could be arranged in parallel between two heat baths such that heat is spontaneously transferred from the cold bath to the hot bath. (b) This is incorrect and is not a statement of the Second Law. Rather, a version of the Second Law as put forward by Kelvin (W. Thomson) is that no engine operating in a cycle can accept a given quantity of heat from one source and convert it solely to work. (c) This is a statement of the second law, although this concept is usually stated as the impossibility of the reverse direction put forward by Clausius that no engine working in a cycle can produce as its only result the transfer of heat from a cold bath to a hot bath. (d) This is incorrect and is not a statement of the Second Law. Rather the entropy of the universe is increasing.

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    Chapter 3 solutions manual to accompany 5’th Edition of Physical Chemistry , Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak 4 4 8 . a) The motor will not work. The ratchet and stick are also hit randomly by gas molecules. Thus, the stick, intended to only allow clockwise rotation of the wheel (ratchet), can be lifted due to collisions with gas molecules at the ratchet. At the moment when it is lifted, there is equal probability that the stick will slip forward or backward one notch due to the random fluctuations of the gas molecules at the paddle. The result is no net motion. b) If the ratchet is at a lower temperature than the paddle, the likelihood that the stick will lift due to fluctuations at the ratchet will be much less (at lower temperature, sufficient energy for this process is less often achieved.) Thus, the lifting of the stick will be determined by the gas molecules hitting the paddle, and the stick will work as intended, only allowing clockwise rotation. In this scenario, the second law is not violated, i.e. heat is not converted to work at constant temperature. 9 . , 0 < q (metabolism converts chemical energy into heat) 0 = Δ T (thermostat); 0 ≅ w (depending on whether gases are produced or consumed) Δ U ≅ q < 0; Δ H ≅ Δ U < 0 ; 0 < Δ G (spontaneous process at constant T, p ) 10 . a) 1kg H 2 = (1000g) (2.016g mol − 1 ) = 496.0mol b) 1kg of n -octane = (1000)/(114.23) = 8.754mol octane Δ f H ° ( H 2 O,1) = ( − 285.83)(496.0) = − 141,780kJ Δ S ° = [ S ° (H 2 O,l) − S ° (H 2 , g) − 0.5 S ° (O 2 ,g)](496 mol) = [69.91 − 130.57 − (0.5)(205.03)](496) = − 80.94kJ K − 1 Δ f G ° = ( − 237.18)(496) = − 117,650 kJ 12.5 O 2 (g) + C 8 H 18 (g) → 8CO 2 (g) + 9H 2 O(g) Δ H ° = [8( − 393.51) + 9( − 285.83) − ( − 208.45)](8.754) = − 48,250 kJ Δ S ° = [8(213.64) + 9(69.91) − 466.73 − (12.5)(205.03)](8.75) = − 6.05 kJ K − 1 Δ G ° = [8( − 394.36) + 9( − 237.18) − 16.40](8.754) = − 46,450 kJ 11 . a) 6C(s, graphite) + 3H 2 (g) → C 6 H 6 (g) Δ r G 298 ° = Δ f G 298 ° (benzene) = 129.86 kJ mol − 1 Don’t buy: non-spontaneous at C 25 ° ,1 bar. Catalysts change rate, not spontaneity. b) 2NO(g) + O 2 (g) → 2NO 2 (g) Δ r G 298 ° = − 2(86.55) + 0 + 2(51.47) = − 70.16 kJ mol − 1 Yes, the reaction is spontaneous. c) 3CH 4 (g) + NH 3 (g) + 3O 2 (g) → CH 3 CHNH 2 COOH(s) + 4H 2 O(l) Δ r G 298 ° = − 3( − 50.70) − ( − 16.78) + 0 + ( − 371.64) + 4( − 237.13) = − 1151.28 kJ mol − 1 Yes, this reaction is spontaneous.

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    Chapter 3 solutions manual to accompany 5’th Edition of Physical Chemistry , Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak 5 5 12 . a) 100 g MgATP 2- * (1 mol/527.42 g) (-50 kJ / mol) = -9.48 kJ b) 6000 kJ / -50 kJ| ( 527.416 g / mol) = 63.3 kg MgATP 2- or 633 turnovers of 100 g c) 6.3 kg or about 14 lbs of MgATP 2- would be needed for this level of storage; this is still excessive, and would be around 10% of a typical body mass; for comparison bone mass percentage is around 15% in most adults 13 . a) irreversibly b) the system plus the surroundings. In this case, the surroundings is a correct answer but the first answer is true for all spontaneous irreversible processes. c) enthalpy change d) less than (consider the analogous process of the Carnot cycle); if the net process is a cycle for the system then entropy of surroundings must have net increase 14 . C 2 H 5 OH(l) → C 2 H 6 (g) + 1 2O 2 (g) Δ r H 298 ° = − ( − 276.98) + ( − 84.0) + 0 = 192.98 kJ mol − 1 Δ r G 298 ° = − ( − 174.09) + ( − 32.08) + 0 = 142.01 kJ mol − 1 Δ r U 298 ° = Δ r H 298 ° − Δ ( pV ) ≅ Δ r H 298 ° − ( Δ n gases ) RT (assuming ideal gas behavior) = 192.98 kJ mol − 1 − (3 2 )(8.314 1000 )(298) = 189.3 kJ mol − 1 To obtain Δ H ο at 500 ο C and 1 bar we would need heat capacity values for each of the reactants and products over that temperature range. To be realistic, we should include the enthalpy of vaporization of ethanol, which is a gas at 500 ο C and 1 bar. Δ r H 773 ° = Δ r H 298 ° + ( Δ C p ) 298 773 ∫ dT + Δ vap H ° 773 (C 2 H 5 OH) 15 . aragonite) ( CaCO calcite) ( CaCO 3 3 + a) Δ G 298 ° = − ( − 1128.79) + ( − 1127.71) = + 1.08 kJ mol − 1 No. The positive Δ G ° says that the process is non-spontaneous. b) Because the density of aragonite is greater than that of calcite, an increase in pressure should favor the conversion to aragonite, according to Le Chatelier’s Principle. c) Δ r G 298 ( p ) − Δ r G 298 ° = Δ r V ( p − 1 bar), where Δ r G 298 ( p ) < 0 (Eq. 3.45)

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    Chapter 3 solutions manual to accompany 5’th Edition of Physical Chemistry , Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak 6 6 p = − 1,050 J mol − 1 [(100.09 g mol − 1 )(1 2.930 − 1 2.710 ) cm 3 g − 1 ] − 1 × (83.14 cm 3 bar deg − 1 8.314 J deg − 1 ) + 1 bar = 3790 bar d) Δ S 2 9 8 ° = 88.70 − 92.88 = − 4.18 J K − 1 mol − 1 Because Δ S ° is negative, that means that Δ G ° will increase with increasing temperature, making the conversion even less favorable. 16 . CH 3 COCOOH(l) → CH 3 CHO(g) + CO 2 (g) a) Δ r G 298 ° = − ( − 463.33) + ( − 133.27) + ( − 394.36) = − 64.30 kJ mol − 1 b) Assuming that both CHO CH 3 and 2 CO remain gases at C 25 ° and 100 bar. Δ G = Δ G °+ RT ln([ p 2 (CH 3 CHO) p 1 ][ p 2 (CO 2 ) p 1 ]) = − 64.30 + (8.314 1000)(298) ln (100 × 100) = − 41.48 kJ mol − 1 17 . a) decrease b) remain unchanged c) zero d) entropy change of the surroundings, entropy change of the universe e) decrease f) remain unchanged g) remain unchanged h) more negative than 18 . a) w = − p φ ( Δ φ V m ) Δ φ U = q φ + w ; Δ φ H = q φ Δ φ S = q φ T φ ; Δ φ G = 0 b) Δ φ H = q φ + ( C p ,m, β − C p ,m, α )( T ∗ − T m ) Δ φ S = q φ T φ + ( C p ,m, β − C p ,m, α )ln T ∗ T φ

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