Chem 101 Book Answers (Ch.9)

Chem 101 Book Answers (Ch.9) - Chapter 9 Reaction...

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Chapter 9 Reaction Energetics 9-1 C H H H N H H O O O C O N N 4 + 9 4 + 10 + 2 H O H 1. What are the signs of Δ H o and Δ S o for the evaporation of water? First write the process: H 2 O(l) H 2 O(g). Evaporation is endothermic ( Δ H o > 0) because the potential energy of the vapor is greater than that of the liquid. The vapor is much more disordered than the liquid so Δ S o > 0. 3. A system gives off 600 J of heat while 200 J of work is done on it. What are Δ E, Δ E sur , and Δ E univ ? q = -600 J and w = 200 J Δ E = q + w = -600 + 200 = -400 J The system gives off 400 J of energy, which goes into the surroundings Δ E sur = - Δ E = +400 J Δ E univ = 0 for any process 5. What is the difference between Δ H and Δ E? Why is Δ H used more frequently? Δ E is the energy change of a system, while Δ H is the heat absorbed by a system at constant temperature and pressure. Most chemical processes are carried out at constant pressure, and the heat absorbed or given off is more important to chemists than the work term in Equation 9.2, so Δ H is used more frequently. 7. When is the entropy change of a reaction expected to be significant? Entropy changes are important when gases are produced or consumed. 9. Indicate whether each of the following is true, false, or cannot be determined for an extensive endothermic reaction that is at equilibrium at constant pressure and temperature. a) Δ G < 0 False. Δ G = 0 at equilibrium. b) Δ H o < 0 False. Reaction is endothermic, so Δ H o > 0 c) Δ G o <0 True. Reaction is extensive. d) Δ S univ > 0 False. Δ S univ > 0 for spontaneous processes, but this process is at equilibiurm, so Δ S univ = 0 e) Δ S o > 0 True. If Δ G o < 0 and Δ H o > 0 then Δ S o must be positive f) the extent of reaction increases with T: True. Δ S o is positive, so increasing T makes Δ G o less positive or more negative. 11. The combustion of 0.150 g of C 2 H 5 OH(l) (grain alcohol) releases 4.47 kJ of heat. a) What is the enthalpy of combustion of C 2 H 5 OH(l)? The heat of combustion is the heat absorbed when one mole of the substance reacts with oxygen, but we are given the amount of heat released when 0.150 g is combusted. The molar mass of C 2 H 5 OH is 46.1 g/mol, and so we may write -4.47 kJ 0.150 g C 2 H 5 OH × 46.1 g C 2 H 5 OH mol C 2 H 5 OH = -1.37 x 10 3 kJ/mol b) Write the chemical equation for the combustion reaction: C 2 H 5 OH(l) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(l) c) What is the enthalpy change for 4CO 2 (g) + 6H 2 O(l) 2C 2 H 5 OH(l) + 6O 2 (g)? The reaction is the reverse of twice the combustion reaction, Δ H o = -2 x Δ H comb o = -2 x (-1.37 x 10 3 ) = +2.74 x 10 3 kJ 13. Use the data in Table 9.2 to estimate the following enthalpies of combustion: a) H 2 C=CH 2 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(g) The oxygen-oxygen bond in O 2 is a double bond as are the two carbon-oxygen bonds in CO 2 . Therefore, four C-H bonds, a C=C bond, and three O=O bonds are broken. Four C=O bonds and four O-H bonds are formed Δ H ~ 4 D C-H + D C=C + 3 D O=O – 4 D C=O – 4 D O-H = 4(413) + 612 + 3(495) – 4(799) – 4(463) = -1299 kJ b) 4CH 3 NH 2 (g) + 9O 2 (g) 4CO 2 (g) + 10H 2 O(g) + 2N 2 (g) Twelve C-H bonds, four C-N, eight N-H and nine O=O bonds must be broken. Eight C=O bonds, twenty O-H bonds and two N
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This note was uploaded on 10/30/2009 for the course CH Ch101 taught by Professor Bigham during the Fall '08 term at N.C. State.

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Chem 101 Book Answers (Ch.9) - Chapter 9 Reaction...

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