Exam 4-solutions-1

# Exam 4-solutions-1 - Version 139 Exam 4 mccord(51600 This...

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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 – fnd all choices beFore answering. mccord -ch301 10am class only - 51600 C s (water) = 4.184 J/g C water’s normal boiling pt = 100 C Thermodynamic Data at 25 C Δ H f Δ 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 CH± 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: Δ H vap = 40500 J · mol 1 T BP = 352 K Δ S cond = q T = Δ H con T BP = Δ 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 Δ 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 Δ 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) p 4 . 184 J g · C P (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 p 34 . 335 kJ g P ± 86 . 1 g mol ² = 2956 kJ / mol

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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. Δ S is positive when there is increasing disorder. 2. w is positive when work is done by the system. correct 3. Δ H is negative when heat is released to the surroundings. 4. Δ G is negative when a reaction is spon- taneous. 5. Work is done on the system when Δ V is negative. Explanation: w is positive only when work is done ON the system, not BY the system. When the system does work, volume increases or the number of moles increases (Δ V > 0, Δ n > 0). w = - P Δ V = - Δ nRT and will therefore be negative when Δ V or Δ n is positive.
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Exam 4-solutions-1 - Version 139 Exam 4 mccord(51600 This...

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