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

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Unformatted text preview: 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 H is positive and 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 = (+) , H = (+) and S = (- ) G = H- T S G = (+)- (+)(- ) = (+)- (- ) = (+) + (+) 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. 2C 3 H 8 ( ) + 10O 2 (g) 6CO 2 (g) + 8H 2 O( ) 2. 2C 8 H 18 ( ) + 25O 2 (g) 16CO 2 (g) + 18H 2 O( ) 3. 2C 8 H 18 ( ) + 25O 2 (g) 16CO 2 (g) + 18H 2 O(g) correct 4. 2C 3 H 8 ( ) + 10O 2 (g) 6CO 2 (g) + 8H 2 O(g) Explanation: The correct answer has the greatest 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 S sys , respectively? 1. +, +, + 2.- , +,- correct 3. +,- ,- 4.- , +, + 5. +, +,- 6. +,- , + Explanation: 004 (part 1 of 1) 10 points Which statement is FALSE? 1. H is sometimes exactly equal to E . 2. H is equal to E for the process 2H 2 (g) + O 2 (g) 2H 2 O(g). correct 3. H is often nearly equal to E . 4. The thermodynamic quantity most easily measured in a coffee cup calorimeter is H . 5. No work is done in a reaction occurring in a bomb calorimeter. Explanation: For 2H 2 (g) + O 2 (g) 2H 2 O(g) , n i = 3 mol gas n f = 2 mol gas n =- 1 mol gas Iype, Shelby Exam 4 Due: Dec 6 2007, 11:00 pm Inst: Vandenbout 2 w =- P V =- ( n ) R T 6 = 0 , Since E = q + w , q = H and w 6 = 0, E 6 = 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 / molK. Explanation: S = H T T = 136 + 273.15 = 409 . 15 K S = 48200 / 409 . 15 = 117 . 805 J / molK 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 ; H f H(g) = 217 . 9 kJ mol- 1 ; 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....
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exam 4 answers - Iype, Shelby Exam 4 Due: Dec 6 2007, 11:00...

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