zumdahl_chemprin_6e_csm_ch09

# zumdahl_chemprin_6e_csm_ch09 - CHAPTER 9 ENERGY, ENTHALPY,...

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341 CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY The Nature of Energy 15. Ball A: PE = mgz = 2.00 kg × 2 s m 81 . 9 × 10.0 m = 2 2 s m kg 196 = 196 J At point I: All 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 × 2 s m 81 . 9 × 3.00 m = 118 J KE = E total PE = 196 J 118 J = 78 J 16. Plot a represents an exothermic reaction. In an exothermic process, the bonds in the product molecules are stronger (on average) than those in the reactant molecules. The net result is that the quantity of energy Δ (PE) is transferred to the surroundings as heat when reactants are converted to products. For an endothermic process (plot b), energy flows into the system as heat to increase the potential energy of the system. In an endothermic process, the products have higher potential energy (weaker bonds on average) than the reactants. 17. Path-dependent functions for a trip from Chicago to Denver are those quantities that depend on the route taken. One can fly directly from Chicago to Denver or one could fly from Chicago to Atlanta to Los Angeles and then to Denver. Some path-dependent quantities are miles traveled, fuel consumption of the airplane, time traveling, airplane snacks eaten, etc. State functions are path independent; they only depend on the initial and final states. Some state functions for an airplane trip from Chicago to Denver would be longitude change, latitude change, elevation change, and overall time zone change. 18. Only when there is a volume change can PV work be done. In pathway I (steps 1 + 2), only the first step does PV work (step 2 has a constant volume of 30.0 L). In pathway II (steps 3 + 4), only step 4 does PV work (step 3 has a constant volume of 10.0 L).

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342 CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY Pathway I: w = ! P Δ V = ! 2.00 atm(30.0 L ! 10.0 L) = -40.0 L atm × atm L J 3 . 101 = ! 4.05 × 10 3 J Pathway II: w = ! P Δ V = ! 1.00 atm(30.0 L ! 10.0 L) = ! 20.0 L atm × atm L J 3 . 101 = ! 2.03 × 10 3 J Note : The sign is minus ( ! ) because the system is doing work on the surroundings (an expansion). We get different values of work for the two pathways; both pathways have the same initial and final states. Because w depends on the pathway, work cannot be a state function. 19. Step 1: Δ E 1 = q + w = 72 J + 35 J = 107 J; step 2: Δ E 2 = 35 J 72 J = 37 J Δ E overall = Δ E 1 + Δ E 2 = 107 J 37 J = 70. J 20. a. Δ E = q + w = ! 23 J + 100. J = 77 J b. w = ! P Δ V = ! 1.90 atm(2.80 L ! 8.30 L) = 10.5 L atm × atm L J 3 . 101 = 1060 J Δ E = q + w = 350. J + 1060 = 1410 J c. w = ! P Δ V = ! 1.00 atm(29.1 L ! 11.2 L) = ! 17.9 L atm × atm L J 3 . 101 = ! 1810 J Δ E = q + w = 1037 J ! 1810 J = ! 770 J 21. q = molar heat capacity × mol × Δ T = mol C J 8 . 20 o × 39.1 mol × (38.0 0.0)°C = 30,900 J = 3 0 . 9 k J w = P Δ V = 1.00 atm × (998 L 876 L) = 122 L atm × atm L J 3 . 101 = 12,400 J = 12.4 kJ Δ E = q + w = 30.9 kJ + ( 12.4 kJ) = 18.5 kJ 22. In this problem, q = w = 950. J. 950. J × J 3 . 101 atm L 1 = 9.38 L atm of work done by the gases w = P Δ 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 23. H 2 O(g) H 2 O(l); Δ E = q + w; q = 40.66 kJ; w = P Δ V
CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 343 Volume of 1 mol H 2 O(l) = 1.000 mol H 2 O(l) × g 996 . 0

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## This note was uploaded on 03/29/2011 for the course CHEM 232 taught by Professor Malambri,w during the Spring '11 term at Kentucky.

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zumdahl_chemprin_6e_csm_ch09 - CHAPTER 9 ENERGY, ENTHALPY,...

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