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Exam 4 Quy

# Exam 4 Quy - Version 028 Exam 4 quy(50970 This print-out...

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Version 028 – Exam 4 – quy – (50970) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Which would you expect to be least viscous? 1. C 8 H 18 at 50 C 2. C 8 H 18 at 30 C 3. C 4 H 8 at 50 C correct 4. C 4 H 8 at 30 C Explanation: 002 10.0points The energy change Δ H associated with the reaction NBr 3 (g) + 3 H 2 O(g) 3 HOBr(g) + NH 3 (g) is +81 kJ/mol rxn. These bond energy values might be useful: O-H 459 kJ/mol; N-H 386 kJ/mol; O-Br 201 kJ/mol. The strength of the N Br bond is 1. 66 kJ/mol 2. 4 kJ/mol 3. 280 kJ/mol 4. 248 kJ/mol 5. 128 kJ/mol 6. 155 kJ/mol correct 7. 465 kJ/mol Explanation: Δ H = 81 kJ Bonds broken: 3 N Br bonds at x kJ each 6 O H bonds at 459 kJ each Bonds made: 3 N H bonds at 386 kJ each 3 O H bonds at 459 kJ each 3 O Br bonds at 201 kJ each. Δ H = (bonds broken) - (bonds made) 81 = [3 x + 6(459)] - [3(386) + 3(459) + 3(201)] 81 = 3 x + 2754 - 3138 3 x = 465 x = 155 kJ 003 10.0points The normal melting point for ice is 273 K. Ice melting at atmospheric pressure and 272 K is 1. exothermic and spontaneous. 2. endothermic and nonspontaneous. cor- rect 3. endothermic and spontaneous. 4. exothermic and nonspontaneous. Explanation: To melt ice, energy must be added, so the process is endothermic and Δ H is positive. At phase change points, Δ G = 0. Below 273 K Δ G will be negative and this will be a nonspontaneous process. 004 10.0points Consider the equation NH 4 Br(s) NH 3 (g) + HBr(g) carefully, and think about the sign of Δ S for the reaction it describes. Δ H = +188.3 kJ. Which response describes the thermodynamic spontaneity of the reaction? 1. We cannot tell from the information given. 2. The reaction is spontaneous only at rela- tively low temperatures. 3. The reaction is spontaneous only at rela- tively high temperatures. correct

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Version 028 – Exam 4 – quy – (50970) 2 4. The reaction is not spontaneous at any temperatures. 5. The reaction is spontaneous at all tem- peratures. Explanation: Entropy ( S ) is high for systems with high degrees of freedom, disorder or randomness and low for systems with low degrees of free- dom, disorder or randomness. S(g) > S ( ) > S (s) . A reaction is spontaneous only when Δ G is negative. Δ H is positive for this reaction and Δ S is positive. Δ G = Δ H - T Δ S = (+) - T (+) = (+) - T Δ G will be negative (spontaneous reactions) only at high values of T . 005 10.0points From the following data (at 298 K), Substance Δ H 0 f S 0 kcal/mol cal/mol · K NO 22.0 50.0 H 2 O - 68 . 0 17.0 NH 3 - 11 . 0 46.0 O 2 0.0 49.0 find Δ G 0 for the reaction 4 NO + 6 H 2 O 4 NH 3 + 5 O 2 . The answer is closest to 1. +200 kcal/mol. 2. +150 kcal/mol. 3. +100 kcal/mol. 4. +250 kcal/mol. correct 5. +300 kcal/mol. Explanation: Δ H 0 rxn = summationdisplay n Δ H f , prod - summationdisplay n Δ H f , rct = [4( - 11 . 0 kcal / mol)] - [4(22 . 0 kcal / mol) + 6( - 68 . 0 kcal / mol)] = 276 kcal / mol Δ S 0 rxn = summationdisplay n S prod - summationdisplay n S rct = [5(49 cal / mol · K) + 4(46 cal / mol · K)] - [4(50 cal / mol · K) + 6(17 cal / mol · K)] = 127 cal / mol · K = 0 . 127 kcal / mol · K Δ G 0 = Δ H 0 - T Δ S 0 = 276 kcal / mol - (298 K) (0 . 127 kcal / mol · K) = 238 . 154 kcal / mol 250 kcal / mol 006 10.0points As a liquid is heated, its vapor pressure 1. disappears.
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Exam 4 Quy - Version 028 Exam 4 quy(50970 This print-out...

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