Exam 4 CH301 Mcord Sp 2009

# Exam 4 CH301 Mcord Sp 2009 - Version 370 Exam 4...

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Version 370 – 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 At the normal boiling point of water, Δ H vap = 40 kJ/mole. What is the entropy change for H 2 O( ) H 2 O(g) ? 1. 400 J/mol · K 2. - 40 kJ/mol · K 3. - 107 J/mol · K 4. 40 kJ/mol · K 5. 107 kJ/mol · K 6. 107 J/mol · K correct Explanation: 002 10.0 points For which reactions I) O 2 (g) + H 2 (g) H 2 O 2 ( ) II) C(s , diamond) + O 2 (g) CO 2 (g) III) N 2 ( ) + 3 F 2 (g) 2 NF 3 ( ) IV) C(s , graphite) + 3 2 O 2 (g) + H 2 (g) CO 2 (g) + H 2 O(g) V) 2 Fe(s) + 3 2 O 2 (g) Fe 2 O 3 (s) would Δ H r = Δ H f ? 1. I, III and V only 2. I, III, IV and V only 3. I, II, III, IV and V 4. I and V only correct 5. II and IV only 6. II, III and IV only 7. IV and V only 8. I and II only Explanation: 003 10.0 points Which of the statements below concerning thermodynamic sign convention is NOT true: 1. Δ H is negative when heat is released to the surroundings. 2. Δ G is negative when a reaction is spon- taneous. 3. Work is done on the system when Δ V is negative. 4. w is positive when work is done by the system. correct 5. Δ S is positive when there is increasing disorder. Explanation: w is positive only when work is done ON the system, not BY the system. When the system does work, volume increases or the number of moles increases (Δ V > 0, Δ n > 0). w = - P Δ V = - Δ nRT and will therefore be negative when Δ V or Δ n is positive. 004 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.

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Version 370 – Exam 4 – McCord – (53745) 2 005 10.0 points For the reaction 2 C(s) + 2 H 2 (g) C 2 H 4 (g) Δ H r = +52 . 3 kJ · mol 1 and Δ S r = - 53 . 07 J · K 1 · mol 1 at 298 K. This reac- tion will be spontaneous at 1. all temperatures. 2. temperatures below 985 K. 3. temperatures below 1015 K. 4. temperatures above 985 K. 5. no temperature. correct Explanation: Δ G = Δ H - T Δ S is used to predict spon- taneity. G is negative for a spontaneous reaction.) T is always positive. We thus have Δ G = (+) - T ( - ). For Δ G to be neg- ative, T would have to be negative, which is not physically possible, so the reaction is nonspontaneous at any temperature. 006 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. you must be thinking of the third law. 2. while the entropy involved in the plant growth decreased, the entropy in the universe increased as a result of this process. correct 3. the plant is not at equilibrium and hence the argument is invalid.
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