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Exam_4 - Version 401 Exam 4 McCord(53110 This print-out...

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Version 401 – Exam 4 – McCord – (53110) 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 CH301mwf This exam is only for McCord’s MWF CH301 class. PLEASE carefully bubble in your UTEID and Version Number! You’ll need some (not all) of the values from this table on the exam. Δ H f Δ G f S Substance ( kJ mol ) ( kJ mol ) ( J mol · K ) CO(g) - 110 . 53 - 137 . 17 197.67 CO 2 (g) - 393 . 51 - 394 . 36 213.74 SO 2 (g) - 296 . 83 - 300 . 19 248.22 H 2 O(l) - 285 . 83 - 237 . 13 69.91 H 2 O(g) - 241 . 82 - 228 . 57 188.83 H 2 S(g) - 20 . 63 - 33 . 56 205.79 CH 3 OH(l) - 238 . 86 - 166 . 27 126.8 CH 4 (g) - 74 . 81 - 50 . 72 186.26 C 2 H 4 (g) +52 . 26 +68 . 12 219.56 C 3 H 8 (g) - 103 . 85 - 23 . 49 270.2 C 4 H 10 (g) - 126 . 15 - 17 . 03 310.1 C 6 H 6 ( ) +49 . 0 +124 . 5 173.3 C 8 H 18 ( ) - 249 . 9 +6 . 4 358 001 10.0 points Calculate the change in entropy when 2.00 moles of ozone are compressed isothermally to one quarter of its original volume. Treat ozone as an ideal gas. 1. - 1.39 J · K 1 2. +23.1 J · K 1 3. - 23.1 J · K 1 correct 4. +10.0 J · K 1 5. - 10.0 J · K 1 Explanation: If compressed to one quarter the original volume this means V 2 = V 1 4 or V 2 V 1 = 1 4 Δ S = nR ln parenleftbigg V 2 V 1 parenrightbigg = (2 . 00 mol)(8 . 314 J · mol 1 · K 1 ) × ln parenleftbigg 1 4 parenrightbigg = - 23 . 0513 J · K 1 We expect a negative answer since the volume decreased. 002 10.0 points Calculate the standard reaction enthalpy for the reaction N 2 H 4 ( ) + H 2 (g) 2 NH 3 (g) given N 2 H 4 ( ) + O 2 (g) N 2 (g) + 2H 2 O(g) Δ H = - 543 kJ · mol 1 2 H 2 (g) + O 2 (g) 2 H 2 O(g) Δ H = - 484 kJ · mol 1 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Δ H = - 92 . 2 kJ · mol 1 1. - 243 kJ · mol 1 2. - 151 kJ · mol 1 correct 3. - 59 kJ · mol 1 4. - 1119 kJ · mol 1 5. - 935 kJ · mol 1 Explanation: We need to reverse the second reaction and add them:
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Version 401 – Exam 4 – McCord – (53110) 2 N 2 H 4 ( ) + O 2 (g) N 2 (g) + 2 H 2 O(g) Δ H = - 543 kJ/mol 2 H 2 O(g) 2 H 2 (g) + O 2 (g) Δ H = +484 kJ/mol N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Δ H = - 92 . 2 kJ/mol N 2 H 4 ( ) + 3 H 2 (g) 2 H 2 (g) + 2 NH 3 (g) N 2 H 4 ( ) + H 2 (g) 2 NH 3 (g) Δ H = - 151 . 2 kJ/mol 003 10.0 points Which one of the following reactions has a positive entropy change? 1. BF 3 (g) + NH 3 (g) F 3 BNH 3 (s) 2. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 3. 2 NH 4 NO 3 (s) 2 N 2 (g) + 4 H 2 O(g) + O 2 (g) correct 4. H 2 O(g) H 2 O( ) 5. 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) Explanation: Processes that result in an increase in the number of moles of gaseous substances have ΔS sys > 0. Only this choice demonstrates an increase in the number of moles of gaseous substances (from 0 moles of gaseous reactants to 7 moles of gaseous products). 004 10.0 points The entropy of fusion of a substance is always smaller than its entropy of vaporization. 1. True correct 2. False Explanation: The entropy of fusion is the entropy change upon melting a solid into a liquid. The en- tropy change of vaporization is the entropy change when a liquid becomes a gas. The en- tropy of a gas is significantly greater than that of a liquid, which is a little more than that of a solid. This is because in a gas the particles are able to move around much more.
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