exam 4 answers - Iype Shelby Exam 4 Due Dec 6 2007 11:00 pm...

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Iype, Shelby – Exam 4 – Due: Dec 6 2007, 11:00 pm – Inst: Vandenbout 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Vanden Bout ONLY ! ! ! ! 001 (part 1 of 1) 10 points A reaction for which Δ H is positive and Δ S is negative 1. could become spontaneous at low temper- atures. 2. is not spontaneous at any temperature. correct 3. is spontaneous at any temperature. 4. could become spontaneous at high tem- peratures. Explanation: T = (+) , Δ H = (+) and Δ S = ( - ) Δ G = Δ H - T Δ S Δ G = (+) - (+)( - ) = (+) - ( - ) = (+) + (+) Δ G will always be positive, so the process will always be nonspontaneous. 002 (part 1 of 1) 10 points For which of the following reactions would you predict the most positive entropy change? 1. 2 C 3 H 8 ( ) + 10 O 2 (g) 6 CO 2 (g) + 8 H 2 O( ) 2. 2 C 8 H 18 ( ) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O( ) 3. 2 C 8 H 18 ( ) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O(g) correct 4. 2 C 3 H 8 ( ) + 10 O 2 (g) 6 CO 2 (g) + 8 H 2 O(g) Explanation: The correct answer has the greatest Δ n gas value. Generating gases from liquids and solids generally increases entropy. 003 (part 1 of 1) 10 points When water condenses, what are the signs for q , w , and Δ S sys , respectively? 1. +, +, + 2. - , +, - correct 3. +, - , - 4. - , +, + 5. +, +, - 6. +, - , + Explanation: 004 (part 1 of 1) 10 points Which statement is FALSE? 1. Δ H is sometimes exactly equal to Δ E . 2. Δ H is equal to Δ E for the process 2 H 2 (g) + O 2 (g) 2 H 2 O(g). correct 3. Δ H is often nearly equal to Δ E . 4. The thermodynamic quantity most easily measured in a coffee cup calorimeter is Δ H . 5. No work is done in a reaction occurring in a bomb calorimeter. Explanation: For 2 H 2 (g) + O 2 (g) 2 H 2 O(g) , n i = 3 mol gas n f = 2 mol gas Δ n = - 1 mol gas
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Iype, Shelby – Exam 4 – Due: Dec 6 2007, 11:00 pm – Inst: Vandenbout 2 w = - P Δ V = - n ) R T 6 = 0 , Since Δ E = q + w , q = Δ H and w 6 = 0, Δ E 6 = Δ H . 005 (part 1 of 1) 10 points Calculate the entropy of vaporization for com- pound X at its boiling point of 136 C. The enthalpy of vaporization of compound X is 48 . 2 kJ/mol. Correct answer: 117 . 805 J / mol K. Explanation: Δ S = Δ H T T = 136 + 273.15 = 409 . 15 K Δ S = 48200 / 409 . 15 = 117 . 805 J / mol K 006 (part 1 of 1) 10 points Find 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. - 1170 kJ · mol - 1 2. - 83 kJ · mol - 1 3. - 44 kJ · mol - 1 correct 4. - 516 kJ · mol - 1 5. - 691 kJ · mol - 1 Explanation: 007 (part 1 of 1) 10 points The First Law of Thermodynamics deals with 1. conservation of mass. 2. conservation of energy. correct 3. creation of energy. 4. destruction of energy. Explanation: The First Law of Thermodynamics states that the total amount of energy in the uni- verse is constant. Energy is conserved; it cannot be created nor destroyed. 008 (part 1 of 1) 10 points 1.95 mol of an ideal gas at 300 K and 3.00 atm expands from 16 L to 28 L and a final pressure of 1.20 atm in two steps: (1) the gas is cooled at constant volume until its pressure has fallen to 1.20 atm, and (2) it is heated and allowed to expand against a constant pressure of 1.20 atm un- til its volume reaches 28 L.
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