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exam 4 solution

exam 4 solution - Version 017 Exam Four Shear Shear(52375...

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Version 017 – Exam Four: Shear – Shear – (52375) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. DrRuth says: In all cases, select the answer that BEST answers the question asked. The total points on this exam is 300, and it counts equally with all previous inclass exams. Read the whole thing first and plan which to answer first because you feel confident about them. Notice the harder ones and save them up till last so you don’t use up all your time working on them. Shear 52375 1 L atm = 101.325 J 1 cal = 4.184 J R = see back of scantron sheet 001 15.0 points When 0.100 g of graphite is burned com- pletely in a bomb calorimeter (heat capacity = 3.344 kJ/ C), containing 3000 g of water, a temperature rise of 0.21 C is observed. What is Δ E for the combustion of graphite? The specific heat of liquid water is 4.184 J/g · C. 1. Δ E = - 285 . kJ/mol 2. Δ E = +3 . 34 kJ/mol 3. Δ E = - 401 . 0 kJ/mol correct 4. Δ E = - 40 . 1 kJ/mol 5. Δ E = - 3 . 34 kJ/mol Explanation: m graphite = 0.100 g m water = 3000 g Δ T = 0.21 C SH water = 4.184 J/g · C HC = 3.344 kJ/ C The amount of heat responsible for the tem- perature increase for 3000 g of water is q = parenleftBig 4 . 184 J g · C parenrightBig (3000 g)(0 . 21 C) parenleftBig 1 kJ 1000 J parenrightBig = 2 . 6359 kJ The amount of heat responsible for the warming of the calorimeter is q = parenleftBig 3 . 344 kJ C parenrightBig (0 . 21 C) = 0 . 7022 kJ The amount of heat released for this reaction is 2 . 6359 kJ + 0 . 7022 kJ = 3 . 3381 kJ The reaction was exothermic and there were 0 . 1 g of graphite, so parenleftBig - 3 . 3381 kJ 0 . 1 g parenrightBigparenleftBig 12 g mol parenrightBig = - 401 kJ mol 002 10.0 points For a given reaction Δ G is very negative. Which of the following is FALSE? 1. The reaction is termed spontaneous. 2. If Δ S is positive, then the reaction will probably be more spontaneous at higher tem- peratures. 3. Δ S for the reaction may be positive or negative. 4. The reaction should occur very rapidly because Δ G is very negative. correct 5. The entropy of the universe must increase as a result of this reaction occuring. Explanation: Thermodynamics determines if a reaction will happen; kinetics determines how fast a reaction will occur. Δ G is a thermodynamic term, not a kinet- ics term. 003 15.0 points For the reaction 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) Δ H r = +198 kJ · mol 1 and Δ S r = 190 J · K 1 · mol 1 at 298 K. The forward reaction will be spontaneous at

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Version 017 – Exam Four: Shear – Shear – (52375) 2 1. temperatures above 1315 K. 2. all temperatures. 3. temperatures above 1042 K. correct 4. temperatures below 1042 K. 5. no temperature. Explanation: Δ H r = +198 kJ / mol Δ S r = 0 . 019 kJ mol · K Δ G 0 = Δ H 0 - T Δ S Δ G 0 < 0 for a spontaneous reaction, so 0 > Δ H 0 - T Δ S T > Δ H 0 Δ S 0 = 198 kJ / mol 0 . 019 kJ mol · K = 1042 . 11 K Thus the temperature would need to be > 1042 . 11 K.
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exam 4 solution - Version 017 Exam Four Shear Shear(52375...

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