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Unformatted text preview: Version 108 Exam 4 mccord (50970) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points In a system at equilibrium at constant tem perature and pressure, H is necessarily equal to 1. T S. correct 2. 0. 3. P V. 4. U. 5. G. Explanation: G = 0 for a system at equilibrium, so G = H T S 0 = H T S H = T S 002 10.0 points The standard molar enthalpy of formation of ammonium chloride, NH 4 Cl (s), is 314 . 43 kJ/mol. What is the standard molar internal energy of formation of ammonium chloride? 1. 314 . 43 kJ/mol 2. 309 . 47 kJ/mol 3. 306 . 99 kJ/mol correct 4. 319 . 39 kJ/mol 5. 321 . 87 kJ/mol 6. +311 . 95 kJ/mol 7. 316 . 91 kJ/mol Explanation: H = 314 . 43 kJ/mol T = 298 . 15 K (standard temperature) 1 2 N 2 (g) + 2 H 2 (g) + 1 2 Cl 2 (g) NH 4 Cl(s) n i = ( . 5 + 2 + . 5)mol = 3 mol n f = 0 mol n = (0 3) mol = 3 mol E = q + w, q = H and w = P V = nRT = ( 3 mol)(8 . 314 J / mol K) (298 . 15 K) parenleftbigg kJ 1000 J parenrightbigg = 7 . 43646 kJ Work is per 1 mole of N 2 H 4 formed, so E = H + w = 314 . 43 kJ / mol + 7 . 43646 kJ / mol = 306 . 994 kJ / mol 003 10.0 points Reactions with positive values of S r always become spontaneous at low temperatures. 1. True 2. False correct Explanation: G = H T S is used to predict spon taneity. ( G is negative for a spontaneous reaction.) T is always positive. When S is positive and T is small, then in order for G to be negative, H must be negative, zero or have a very small positive value. But this is not always the case, so not ALL reactions with a positive S are spontaneous at low temperatures. 004 10.0 points Calculate the standard entropy of condensa tion of chloroform at its boiling point (335 K). The standard molar enthalpy of vaporiza tion of chloroform at its boiling point is 31.4 kJ mol 1 . Version 108 Exam 4 mccord (50970) 2 1. +506 J K 1 mol 1 2. +93 . 7 J K 1 mol 1 3. +31 . 4 kJ K 1 mol 1 4. 93 . 7 J K 1 mol 1 correct 5. 31 . 3 kJ K 1 mol 1 Explanation: H vap = 31 . 4 kJ mol 1 = 31400 J mol 1 T BP = 335 K S cond = q T = H con T BP = H vap T BP = 31400 J mol 1 335 K = 93 . 7313 J mol 1 K 1 005 10.0 points A can of spray paint uses isobutane (C 4 H 10 ) as a propellant. The brand new can contains 15 g of propellant but after several uses the can now contains 13 . 5 g of propellant. What amount of work has been done by the propel lant gas as it expands out of the can? Assume the can is stored and operated at 24 C....
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 Fall '07
 Fakhreddine/Lyon
 Chemistry, Equilibrium

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