Ex_3A_KEY.pdf

# 3612 o 2 g 2050 co 2 g 3935 2136 h 2 o 2858 6991 6

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) ?? 361.2 O 2 (g) 0 205.0 CO 2 (g) –393.5 213.6 H 2 O( ! ) –285.8 69.91 (6 pts) (a). Find the missing D H f ° value for C 8 H 18 ( ! ). D H ° = 16 D H f ° (CO 2 ) + 18 D H f ° (H 2 O) – 2 D H f ° (C 8 H 18 ) – 25 D H f ° (O 2 ) = –10,939.8 kJ 2 D H f ° (C 8 H 18 ) =16 D H f ° (CO 2 ) + 18 D H f ° (H 2 O) – 25 D H f ° (O 2 ) + 10,939.8 kJ = 16 mol (–393.5 kJ/mol) + 18 mol(–285.8 kJ/mol) – 25(0) +10,939.8 kJ = –500.6 kJ D H f ° (C 8 H 18 ) = –500.6 kJ/2 mol = –250.3 kJ/mol (6 pts) (b). Find D S ° for the combustion of octane. D S ° = 16 S ° (CO 2 ) + 18 S ° (H 2 O) – 2 S ° (C 8 H 18 ) – 25 S ° (O 2 ) =16 mol (213.6 J/mol/K) + 18 mol(69.91 J/mol/K) – 2 mol(361.2 J/mol/K) – 25 mol(205.0 J/mol/K) = –1171 J/K

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A8 © 2018 L.S. Brown (Problem 19, continued) 2 C 8 H 18 ( ! ) + 25 O 2 (g) ® 16 CO 2 (g) + 18 H 2 O( ! ) D H ° = –10,939.8 kJ (8 pts) (c). Calculate the amount of heat that would be produced by the complete combustion of 1.0 gallon of octane. (The density of liquid octane is 0.69 g/mL, and 1 gallon = 3.785 L.) 1 gal × 3785 mL 1 gal × 0.69 g 1 mL × 1 mol C 8 H 18 114.224 = 22.8 mol The reaction as written has 2 moles of C 8 H 18 , so we would release 10,939.8 kJ for the combustion of 2 moles. Scale that up to the 22.8 moles: ࠵?࠵?. ࠵? ࠵?࠵?࠵? ࠵? ࠵?࠵?, ࠵?࠵?࠵?. ࠵? ࠵?࠵? ࠵? ࠵?࠵?࠵? = ࠵?. ࠵?࠵? × ࠵?࠵? ࠵? ࠵?࠵?
NAME:_________________________________ © 2018 L.S. Brown A9 (12 pts) 20. You make some iced tea by dropping 325 grams of ice into 500.0 mL of warm tea in an insulated pitcher. If the tea is initially at 30.0 ° C and the ice cubes are initially at 0.0 ° C, how many grams of ice will still be present when the contents of the pitcher reach a final temperature? The tea is mostly water, so assume that it has the same density (1.0 g/mL), molar mass, heat capacity (75.3 J/K/mol), and heat of fusion (6.01 kJ/mol) as pure water. The heat capacity of ice is 37.7 J/K/mol. First thing here is to see that final T must be 0 ° C if there is to be any ice left over. And if that’s true, then there is no temperature change for the ice, only melting. So: q ice = –q tea n melt D H fusion = –n tea c tea D T In this, n melt is the amount of ice that has actually melted, NOT the initial ice!
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• Fall '07
• GENERALCHEMFORENG
• Chemistry, pts, Dr. Larry Brown, L.S. Brown

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