Exam 4-solutions-1 - Version 139 Exam 4 mccord (51600) 1...

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Unformatted text preview: Version 139 Exam 4 mccord (51600) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. mccord-ch301 10am class only - 51600 C s (water) = 4.184 J/g C waters normal boiling pt = 100 C Thermodynamic Data at 25 C H f G f C p ( kJ mol ) ( kJ mol ) ( J mol K ) H 2 O(l)-285.8-237.2 75.3 CCl 4 (l)-135.4-65.21 131.75 CHF 3 (g)-695.4 51.0 001 4.0 points Calculate the standard entropy of vaporiza- tion of ethanol at its boiling point 352 K. The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ mol 1 . 1.- 115 J K 1 mol 1 2. +40.5 kJ K 1 mol 1 3. +513 J K 1 mol 1 4.- 40.5 kJ K 1 mol 1 5. +115 J K 1 mol 1 correct Explanation: H vap = 40500 J mol 1 T BP = 352 K S cond = q T = H con T BP = H vap T BP = 40500 J mol 1 352 K = +115 . 057 J mol 1 K 1 002 5.0 points A 2.00 gram sample of a hypothetical sub- stance with molecular weight of 86.1 g/mol undergoes complete combustion with excess O 2 in a bomb calorimeter. The tempera- ture of the 1502 g of water surrounding the bomb rises from 22.65 C to 29.30 C. The heat capacity of the hardware component of the calorimeter (everything that is not water) is 4042 J/ C. What is U for the combustion of this substance? 1.- 3 . 70 10 4 kJ/mol 2.- 5 . 75 10 3 kJ/mol 3.- 2 . 96 10 3 kJ/mol correct 4.- 2 . 26 10 3 kJ/mol 5.- 4 . 97 10 4 kJ/mol Explanation: m C 6 H 8 = 2.00 g m water = 1502 g SH = 4.184 J/g C HC = 4042 J/ C T = 29 . 30 C- 22 . 64 C = 6 . 65 C The increase in the water temperature is 29.30 C- 22.64 C = 6.65 C. The amount of heat responsible for this increase in tempera- ture for 1502 g of water is q = (6 . 66 C) parenleftbigg 4 . 184 J g C parenrightbigg (1502 g) = 41791 J = 41 . 79 kJ The amount of heat responsible for the warm- ing of the calorimeter is q = (6 . 65 C)(4042 J / C) = 26879 J = 26 . 88 kJ The amount of heat released on the reaction is thus 41.79 kJ + 26.88 kJ = 68.67 kJ per the given 2.00 g of n-hexane, which is 34.335 kJ per gram. Per mol of n-hexane, this becomes parenleftbigg 34 . 335 kJ g parenrightbigg parenleftBig 86 . 1 g mol parenrightBig = 2956 kJ / mol Version 139 Exam 4 mccord (51600) 2 However, since heat is released by the system, the sign is negative. 003 4.0 points Which of the statements below concerning thermodynamic sign convention is NOT true: 1. S is positive when there is increasing disorder. 2. w is positive when work is done by the system. correct 3. H is negative when heat is released to the surroundings....
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Exam 4-solutions-1 - Version 139 Exam 4 mccord (51600) 1...

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