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ch301 exam 4 - Exam 4 mccord(51600 This print-out should...

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– 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 10amclassonly-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 molK ) 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 4.0points 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. +40.5 kJ · K 1 · mol 1 2. - 115 J · 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 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 4.0points In a system at equilibrium at constant tem- perature and pressure, Δ H is necessarily equal to 1. Δ G. 2. P Δ V. 3. T Δ S. 4. Δ U. 5. 0. Explanation: Δ G = 0 for a system at equilibrium, so Δ G = Δ H - T Δ S 0 = Δ H - T Δ S Δ H = T Δ S 003 4.0points An endothermic reaction corresponds to one in which 1. change in energy is positive; heat is ab- sorbed. 2. change in energy is positive; heat is evolved. 3. change in energy is negative; heat is evolved. 4. change in energy is negative; heat is ab- sorbed. Explanation: Endothermic reactions require energy to flow from the surroundings into the system. In other words, heat is absorbed by the sys- tem. Δ H , or the change in energy, is positive for endothermic reactions since the amount of energy in the system increased. 004 5.0points A 2.00 gram sample of a hypothetical sub- stance with molecular weight of 86.1 g/mol -`¡Ìi`ÊÜ¡Ì¢ÊÌ¢iÊ`i£¤ÊÛiÀᤥʤvÊ 6¡v¢ÝÊ*À£Ê*,3Ê-`¢Ì£À Ê /¤ÊÀi£¤ÛiÊÌ¢¡ÃÊ¥¤Ì¡Vi]ÊۡáÌ\Ê ÜÜÜ°¡Vi¥¡°V¤£ÉÕ¥¦¤V§°¢Ì£
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– Exam 4 – mccord – (51600) 2 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. - 5 . 75 × 10 3 kJ/mol 2. - 4 . 97 × 10 4 kJ/mol 3. - 2 . 26 × 10 3 kJ/mol 4. - 2 . 96 × 10 3 kJ/mol 5. - 3 . 70 × 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) 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.
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