CH-Exam 1-HW 1 - toupal (rgt374) Homework 2 Sutcliffe...

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toupal (rgt374) – Homework 2 – Sutclife – (52410) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. This HW is due at 11pm the day be±ore Exam 1. 001 10.0 points Calculate the standard Gibbs ±ree energy at 298 K ±or the reaction 4NH 3 (g) + 7 O 2 (g) 4NO 2 (g) + 6 H 2 O( ± ) Species Δ H 0 f S 0 kJ/mol J/mol · K NH 3 (g) - 46 . 11 192 . 3 O 2 (g) 0 . 02 0 5 . 0 NO 2 (g) +33 . 22 4 0 . 0 H 2 O( ± ) - 285 . 86 9 . 91 1. - 1152 kJ / mol rxn correct 2. - 825 kJ / mol rxn 3. +1643 kJ / mol rxn 4. - 1398 kJ / mol rxn 5. - 1643 kJ / mol rxn 6. +1152 kJ / mol rxn 7. +1398 kJ / mol rxn 8. - 180 , 455 kJ / mol rxn 9. +244 , 363 kJ / mol rxn 10. +825 kJ / mol rxn Explanation: Δ H 0 rxn = ± n Δ H 0 fprod - ± n Δ H 0 frct = ² 6( - 285 . 8kJ / mol) +4(33 . 2kJ / mol) ³ - 4( - 46 . 11 kJ / mol) = - 1397 . 56 kJ / mol Δ S 0 rxn = ± n Δ S 0 - ± n Δ S 0 = ² 6(69 . 91 J / mol · K) +4(240 . 0J / mol · K) ³ - ² 7(205 . / mol · K) +4(192 . 3J / mol · K) ³ = - 824 . 74 J / mol · K = - 0 . 82474 kJ / mol · K Δ G 0 H 0 - T Δ S 0 = - 1397 . 56 kJ / mol - (298 K) ( - 0 . 82474 kJ / mol · K) = - 1151 . 79 kJ / mol 002 10.0 points ²or a given reaction at 300 K, Δ G = - 412 kJ/mol rxn and at 400 K, Δ G = - 439 kJ/mol rxn. The enthalpy ±or the reaction is the same at both temperatures (Δ H = - 331 kJ/mol rxn). What is the Δ S (entropy) ±or this reaction? 1. 270 J/mol K correct 2. 27000 J/mol K 3. 0.0 J/mol K 4. 80000 J/mol K 5. - 27000 J/mol K Explanation: Δ G 1 = - 412 kJ / mol rxn T 1 =300K Δ G 2 = - 439 kJ / mol rxn T 2 =400K Δ G H - T Δ S Δ S = Δ G - Δ H - T and Δ H G + T Δ S
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toupal (rgt374) – Homework 2 – Sutclife – (52410) 2 Δ H 1 H 2 = - 331 kJ / mol rxn Δ S 1 = ( - 412 + 331) kJ / mol rxn - 300 K =0 . 27 kJ / K · mol rxn = 270 J / mol · K Δ S 2 = ( - 439 + 331) kJ / mol rxn - 400 K . 27 kJ / K · mol rxn = 270 J / mol · K Alternate Explanation : Δ G 1 + T 1 Δ S G 2 + T 2 Δ S Δ G 1 - Δ G 2 = T 2 Δ S - T 1 Δ S S ( T 2 - T 1 ) Δ S = Δ G 1 - Δ G 2 T 2 - T 1 = ( - 412 + 439) kJ / mol rxn (400 - 300) K . 27 kJ / K · mol rxn =270J / mol · K 003 10.0 points For the reaction 2SO 3 (g) 2 (g) + O 2 (g) Δ H r =+198kJ · mol - 1 at 298 K. Which state- ment is true ±or this reaction? 1. The reaction will not be spontaneous at high temperatures. 2. The reaction will not be spontaneous at any temperature. 3. Δ G r will be positive at high tempera- tures. 4. The reaction is driven by the enthalpy. 5. Δ G r will be negative at high tempera- tures. correct Explanation: Δ G H - T Δ S is used to predict spon- taneity. (Δ G is negative ±or a spontaneous reaction.) Δ H is positive and T is always pos- itive. For the reaction 2 mol gas 3molgas. The more moles o± gas, the higher the disor- der, so Δ S is positive and Δ G =(+) - T (+). For Δ G to be negative, T must be large. 004 10.0 points For the reaction 2C(s)+2H 2 (g) C 2 H 4 (g) Δ H r =+ 5 2 . 3k J · mol - 1 and Δ S r = - 53 . 07 J · K - 1 · mol - 1 at 298 K. The reverse reaction will be spontaneous at 1. no temperatures. 2. temperatures below 1015 K. 3. temperatures below 985 K. 4. all temperatures. correct 5. temperatures above 985 K. Explanation: Δ G H - T Δ S is used to predict spon- taneity. (Δ G is negative ±or a spontaneous reaction.) T is always positive; ±or the reverse reaction, we reverse the sign o± Δ H and Δ S .
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This note was uploaded on 02/13/2012 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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CH-Exam 1-HW 1 - toupal (rgt374) Homework 2 Sutcliffe...

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