mccord exam 4 (2)

# mccord exam 4 (2) - Version 226 Exam 4 McCord(53745 This...

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Version 226 – Exam 4 – McCord – (53745) 1 This print-out should have 26 questions. Multiple-choice 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 0 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 = n C p , m ln T 2 T 1

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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. How many moles of BF 3 does the container hold? You should treat BF 3 ideally. 1. 6.67 mol 2. 4.00 mol 3. 2.50 mol correct 4. 3.33 mol 5. 2.86 mol 6. 1.50 mol Explanation: C P for a non-linear ideal gas is 4 R . The temperature increase is 75 K. q = n C P Δ T 6240 = n (4 · 8 . 314)(75) n = 2 . 5018 mol 007 10.0 points Which of the statements below concerning thermodynamic sign convention is NOT true: 1. Δ S is positive when there is increasing disorder.
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