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# solution2_pdf - toupal(rgt374 – Homework 2 – Sutcliffe...

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Unformatted text preview: toupal (rgt374) – Homework 2 – Sutcliffe – (52410) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This HW is due at 11pm the day before Exam 1. 001 10.0 points Calculate the standard Gibbs free energy at 298 K for the reaction 4 NH 3 (g) + 7 O 2 (g) → 4 NO 2 (g) + 6 H 2 O( ℓ ) Species Δ H f S kJ/mol J/mol · K NH 3 (g)- 46 . 11 192 . 3 O 2 (g) . 205 . NO 2 (g) +33 . 2 240 . H 2 O( ℓ )- 285 . 8 69 . 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 rxn = summationdisplay n Δ H f prod- summationdisplay n Δ H f rct = bracketleftBig 6(- 285 . 8 kJ / mol) +4(33 . 2 kJ / mol) bracketrightBig- 4(- 46 . 11 kJ / mol) =- 1397 . 56 kJ / mol Δ S rxn = summationdisplay n Δ S f prod- summationdisplay n Δ S f rct = bracketleftBig 6(69 . 91 J / mol · K) +4(240 . 0 J / mol · K) bracketrightBig- bracketleftBig 7(205 . 0 J / mol · K) +4(192 . 3 J / mol · K) bracketrightBig =- 824 . 74 J / mol · K =- . 82474 kJ / mol · K Δ G = Δ H- T Δ S =- 1397 . 56 kJ / mol- (298 K) (- . 82474 kJ / mol · K) =- 1151 . 79 kJ / mol 002 10.0 points For a given reaction at 300 K, Δ G =- 412 kJ/mol rxn and at 400 K, Δ G =- 439 kJ/mol rxn. The enthalpy for the reaction is the same at both temperatures (Δ H =- 331 kJ/mol rxn). What is the Δ S (entropy) for 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 = 300 K Δ G 2 =- 439 kJ / mol rxn T 2 = 400 K Δ G = Δ H- T Δ S Δ S = Δ G- Δ H- T and Δ H = Δ G + T Δ S toupal (rgt374) – Homework 2 – Sutcliffe – (52410) 2 Δ H 1 = Δ 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 = 0 . 27 kJ / K · mol rxn = 270 J / mol · K Alternate Explanation : Δ G 1 + T 1 Δ S = Δ G 2 + T 2 Δ S Δ G 1- Δ G 2 = T 2 Δ S- T 1 Δ S = Δ 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 = 0 . 27 kJ / K · mol rxn = 270 J / mol · K 003 10.0 points For the reaction 2 SO 3 (g) → 2 SO 2 (g) + O 2 (g) Δ H ◦ r = +198 kJ · mol − 1 at 298 K. Which state- ment is true for 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....
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solution2_pdf - toupal(rgt374 – Homework 2 – Sutcliffe...

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