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Unformatted text preview: Version 226 – Exam 4 – McCord – (53745) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 1 cal = 4.184 J 1 L atm = 101.325 J R = 8.314 J/mol K 001 10.0 points The internal energy of a substance can be brought to zero by cooling the substance all the way down to absolute zero ( T = 0 K). 1. False correct 2. True Explanation: The very fact that you have any matter at all means you have some internal energy  even at absolute zero. 002 10.0 points Given the standard heats of reaction Reaction Δ H M(s) + 2 X 2 (g) → MX 4 (g) 73 . 7 kJ/mol X 2 (g) → 2 X(g) +342 . 3 kJ/mol M(g) → M(s) 15 . 1 kJ/mol calculate the average bond energy for a single M X bond. 1. 162.35 2. 173.85 3. 178.85 4. 202.85 5. 215.35 6. 168.35 7. 193.35 8. 171.35 9. 197.85 10. 182.35 Correct answer: 193 . 35 kJ / mol. Explanation: The correct reaction will be 1 4 MX 4 (g) → M(g) + 4 X(g) so you must flip the first and third reactions and double the second reaction. 003 10.0 points A friend says that as a plant grows and takes an unordered set of molecules and puts them in a highly organized system the second law of thermodynamics is violated. Your best response is that 1. the plant is not at equilibrium and hence the argument is invalid. 2. you must be thinking of the third law. 3. the second law applies only to chemical systems and not biological systems. 4. while the entropy involved in the plant growth decreased, the entropy in the universe increased as a result of this process. correct Explanation: The plant became more organized, but the universe became less organized. Chemical reacions within the plant are exergonic ( Δ G ) and drive the plant growth forward. 004 10.0 points The temperature of 2.00 mol CO 2 (g) is in creased from 10 ◦ C to 150 ◦ C at constant pres sure. Calculate the change in the entropy of carbon dioxide. Assume ideal behavior. 1. +23 . 39 J/K correct 2. +11 . 70 J/K 3. 23 . 39 J/K 4. 11 . 70 J/K 5. +17 . 71 J/K Explanation: T 1 = 10 ◦ C = 283 K T 2 = 150 ◦ C = 423 K n = 2 mol R = 8 . 314 J mol · K For a linear ideal gas at constant pressure C p , m = 7 2 R , so Δ S = nC p , m ln T 2 T 1 Version 226 – Exam 4 – McCord – (53745) 2 = n 7 2 R ln T 2 T 1 = (2 mol) 7 2 ( 8 . 314 J mol · K ) × ln 423 K 283 K = 23 . 3912 J / K 005 10.0 points The definition of internal energy is Δ U = q + w . Which of these three values are state func tions? 1. Δ U , q , and w 2. q and w only 3. Δ U , but only at constant volume 4. Δ U only correct 5. q and w , but only at constant volume Explanation: 006 10.0 points The temperature of a container of BF 3 (g) goes from 298 K to 373 K at a constant pres sure of 1.00 atm when 6.24 kJ is added as heat....
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 Fall '07
 Fakhreddine/Lyon
 Thermodynamics, Entropy, mol

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