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# solut09 - CHAPTER NINE ENERGY ENTHALPY AND THERMOCHEMISTRY...

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259 CHAPTER NINE ENERGY, ENTHALPY, AND THERMOCHEMISTRY The Nature of Energy 15. Ball A: PE = mgz = 2.00 kg × s m 9.80 2 × 10.0 m = s m kg 196 2 2 = 196 J At Point I: All of this energy is transferred to Ball B. All of B’s energy is kinetic energy at this point. E total = KE = 196 J. At point II, the sum of the total energy will equal 196 J. At Point II: PE = mgz = 4.00 kg × s m 9.80 2 × 3.00 m = 118 J KE = E total - PE = 196 J - 118 J = 78 J 16. KE = 2 1 mv 2 = 2 1 × 2.0 kg × ÷ ø ö ç è æ s m 1.0 2 = 1.0 J; KE = 2 1 mv 2 = 2 1 × 1.0 kg × ÷ ø ö ç è æ s m 2.0 2 = 2.0 J The 1.0 kg object with a velocity of 2.0 m/s has the greater kinetic energy. 17. q = molar heat capacity × mol × T = mol C J 20.8 ° × 39.1 mol × (38.0 - 0.0) ° C = 30,900 J = 30.9 kJ w = -P V = -1.00 atm × (998 L - 876 L) = -122 L atm × atm L J 101.3 = -12,400 J = -12.4 kJ E = q + w = 30.9 kJ + (-12.4 kJ) = 18.5 kJ 18. In this problem q = w = -950. J -950. J × J 101.3 atm L 1 = -9.38 L atm of work done by the gases.

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260 w = -P V, -9.38 L atm = 760 650. - atm × (V f - 0.040 L), V f - 0.040 = 11.0 L, V f = 11.0 L 19. H 2 O(g) H 2 O(l); E = q + w; q = -40.66 kJ; w = -P V Volume of one mol H 2 O(l) = 1.000 mol H 2 O(l) × g 0.996 cm 1 _ mol g 18.02 3 = 18.1 cm 3 = 18.1 mL w = -P V = - 1.00 atm × (0.0181 L - 30.6 L) = 30.6 L atm × atm L J 101.3 = 3.10 × 10 3 J = 3.10 kJ E = q + w = -40.66 kJ + 3.10 kJ = -37.56 kJ Properties of Enthalpy 20. A state function is a function whose change depends only on the initial and final states and not on how one got from the initial to the final state. If H and E were not state functions, the law of conservation of energy (first law) would not be true. 21. One should try to cool the reaction mixture or provide some means of removing heat since the reaction is very exothermic (heat is released). The H 2 SO 4 (aq) will get very hot and possibly boil, unless cooling is provided. 22. a. 1.00 g CH 4 × CH mol kJ 891 - _ CH g 16.04 CH mol 1 4 4 4 = -55.5 kJ b. PV = nRT, n = K 298 _ K mol atm L 0.08206 L 10 _ 1.00 _ atm 760 740. = RT PV 3 = 39.8 mol 39.8 mol × mol kJ 891 - = -3.55 × 10 4 kJ 23. 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) H = -1652 kJ; Note that 1652 kJ of heat are released when 4 mol Fe reacts with 3 mol O 2 to produce 2 mol Fe 2 O 3 . a. 4.00 mol Fe × Fe mol 4 kJ 1652 - = -1650 kJ heat released
CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 261 b. 1.00 ml Fe 2 O 3 × O Fe mol 2 kJ 1652 - 3 2 = -826 kJ heat released c. 1.00 g Fe × Fe mol 4 kJ 1652 - _ g 55.85 Fe mol 1 = -7.39 kJ heat released d. 10.0 g Fe × g 55.85 Fe mol 1 = 0.179 mol Fe; 2.00 g O 2 × g 32.00 O mol 1 2 = 0.0625 mol O 2 0.179 mol Fe/0.0625 mol O 2 = 2.86; The balanced equation requires a 4 mol Fe/3 mol O 2 = 1.33 mol ratio. O 2 is limiting since the actual mol Fe/mol O 2 ratio is less than the required mol ratio. 0.0625 mol O

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solut09 - CHAPTER NINE ENERGY ENTHALPY AND THERMOCHEMISTRY...

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