Chemistry 301 Exam 4 - Barkley, Thane Exam 4 Due: Dec 6...

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Barkley, Thane – Exam 4 – Due: Dec 6 2007, 11:00 pm – Inst: McCord 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. McCord 11 AM ONLY ! ! ! ! 001 (part 1 oF 1) 10 points The Formation oF solid calcium chloride From a gas oF its ions is an endothermic process. 1. True 2. ±alse correct Explanation: 002 (part 1 oF 1) 10 points A process is thermodynamically reversible iF 1. the fnal and initial states are the same. 2. The total entropy change is zero. correct 3. The heat and work are equal but opposite in sign. 4. the total entropy change is positive. 5. The heat is zero. Explanation: ±or a process to be reversible, it must re- sult in a total (universal) entropy change oF zero. Then neither the Forward nor the re- verse process results in decrease in the total entropy. 003 (part 1 oF 1) 10 points Calculate the entropy oF vaporization For com- pound X at its boiling point oF 115 C. The enthalpy oF vaporization oF compound X is 34 . 3 kJ/mol. Correct answer: 88 . 3679 J / mol K. Explanation: Δ S = Δ H T T = 115 + 273.15 = 388 . 15 K Δ S = 34300 / 388 . 15 = 88 . 3679 J / mol K 004 (part 1 oF 1) 10 points ±ind the standard enthalpy oF Formation For NH 3 (g) given N H bond enthalpy = 390 kJ · mol - 1 ; Δ H f H(g) = 217 . 9 kJ · mol - 1 ; Δ H f N(g) = 472 . 6 kJ · mol - 1 . 1. - 44 kJ · mol - 1 correct 2. - 83 kJ · mol - 1 3. - 1170 kJ · mol - 1 4. - 691 kJ · mol - 1 5. - 516 kJ · mol - 1 Explanation: 005 (part 1 oF 1) 10 points A Friend states that water Freezing is a vio- lation oF the second law oF thermodynamics. Your best reply would be that 1. you must be thinking oF the frst law. 2. although the entropy oF the water de- creases the entropy oF the universe increased. correct 3. the second law does not apply at 0 C. 4. the second law does not apply to water. Explanation: The Second Law oF Thermodynamics says that in spontaneous changes, the universe tends toward a state oF greater disorder. 006 (part 1 oF 1) 10 points Calorimetry can be used to fnd heats oF com- bustion. Suppose a calorimeter helps some- one to determine that 4032 kJ oF heat is re- leased during the combustion oF 88.0 grams oF propane, C 3 H 8 (44.0 g/mol). What would be the amount oF heat released For the combus- tion oF 0.500 moles oF propane?
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Barkley, Thane – Exam 4 – Due: Dec 6 2007, 11:00 pm – Inst: McCord 2 1. 1008 kJ correct 2. 504 kJ 3. 2016 kJ 4. 4032 kJ 5. 8064 kJ Explanation: q = 4032 kJ m = 88.0 g n = 0.500 mol MW = 44.0 g/mol If we know how much heat is evolved from combusting 88 g of propane, then we can use the molecular weight of propane to calculate how much heat is evolved from the combus- tion of 0.5 moles: µ 4032 kJ 88 . 0 g C 3 H 8 ¶µ 44 . 0 g C 3 H 8 1 mol C 3 H 8 × 0 . 5 mol C 3 H 8 = 1008 kJ 007 (part 1 of 1) 10 points For the reaction H 2 (g) + Cl 2 (g) 2 HCl(g) at constant temperature and pressure, we
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Chemistry 301 Exam 4 - Barkley, Thane Exam 4 Due: Dec 6...

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