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# solutions - Version 137 Exam 4 McCord(91750 This print-out...

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Version 137 – Exam 4 – McCord – (91750) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Water data is on the back of your bub- blesheet. 001 10.0 points Estimate the heat released when ethene ( CH 2 CH 2 ) reacts with hydrogen gas to give CH 3 CH 3 . Bond enthalpies are C H : 412 kJ/mol; C C : 348 kJ/mol; C C : 612 kJ/mol; C Br : 276 kJ/mol; H H : 436 kJ/mol. 1. 560 kJ / mol 2. 342 kJ / mol 3. 288 kJ / mol 4. 124 kJ / mol correct 5. 148 kJ / mol Explanation: C H H C H H + H H -→ C H H H C H H H Δ H = summationdisplay E break - summationdisplay E make = bracketleftBig (C C) + (H H) bracketrightBig - bracketleftBig 2 (C H) + (C C) bracketrightBig = (612 kJ / mol + 436 kJ / mol) - bracketleftBig 2 (412 kJ / mol) + 348 kJ / mol bracketrightBig = - 124 kJ / mol , so 124 kJ/mol of heat was released. 002 10.0 points Calculate the change in molar entropy when the pressure of argon is allowed to double isothermally. Assume ideal behavior. 1. +1 . 39 J · mol 1 · K 1 2. +5 . 76 J · mol 1 · K 1 3. - 4 . 16 J · mol 1 · K 1 4. - 1 . 39 J · mol 1 · K 1 5. - 5 . 76 J · mol 1 · K 1 correct Explanation: n = 1 mol P fin = 2 P ini R = 8 . 314 J mol · K The change in molar entropy is Δ S = nR ln parenleftbigg P ini P fin parenrightbigg = nR ln parenleftbigg P ini 2 P ini parenrightbigg = (1 mol) parenleftbigg 8 . 314 J mol · K parenrightbigg ln parenleftbigg 1 2 parenrightbigg = - 5 . 76283 J mol · K 003 10.0 points Which one of the following statements is the best statement of the Second Law of Thermo- dynamics? 1. The absolute entropy of a perfect crystal at 0 K would be 0. 2. For any spontaneous process, the total entropy of the universe must increase. cor- rect 3. For any reversible process, the entropy change of the universe must be positive. 4. For any spontaneous process, the change in entropy of the universe must be negative. 5. For any spontaneous process, the entropy of the system must increase. Explanation: The Second Law of Thermodynamics states that in spontaneous changes, the universe tends toward a state of greater disorder.

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Version 137 – Exam 4 – McCord – (91750) 2 004 10.0 points Which of O 2 (g) , O 2 ( ) , H 2 (g) , H 2 ( ) , H 2 O(g) , H 2 O( ) have a heat of formation equal to zero? 1. O 2 (g), O 2 ( ), H 2 (g), H 2 ( ), H 2 O(g), H 2 O( ) 2. O 2 (g), H 2 (g), H 2 O(g) 3. O 2 (g), H 2 (g) correct 4. O 2 (g), O 2 ( ), H 2 (g), H 2 ( ) 5. All of them, but only at absolute zero Explanation: Molecules in their native state at STP have a heat of formation of zero. 005 10.0 points A friend is into heavy metal. On that note, if he supplied 3900 Joules of heat to a 100 gram chunk of LEAD (a rather heavy metal with a density of 11.34 g/cm 3 ) at 25 C, and the temperature rose to 325 C, what is the specific heat of lead? 1. 0.130 J/g · C correct 2. 0.444 J/g · C 3. 0.120 J/g · C 4. 1.47 J/g · C 5. 0.068 J/g · C Explanation: q = 3900 J m = 100 g T 1 = 25 C T 2 = 325 C density = 11.34 g/cm 3 q = SH · m · Δ T SH = 1 m · Δ T = 3900 J (100 g)(300 C) = 0 . 130 J/g · C 006 10.0 points When a sugar cube dissolves in a cup of coffee (an endothermic process), entropy changes of the sugar plus water, the surroundings, and
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