CHM 132 - Exam 2 - 2006 Key

CHM 132 - Exam 2 - 2006 Key - Chemistry 132 Hour...

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Chemistry 132 Hour Examination 2 March 23, 2006 Name ( PRINT)______ SOLUTIONS _________ Student number____________________________ 0. (5 points) Section ____________________ 1. (30 points ) ____________________ 2. (30 points ) ____________________ 3. (35 points ) ____________________ 4. (30 points ) ____________________ TOTAL (130 points ) “Make your thinking visible.” Show your work neatly and circle your answers . Give correct units for answers requiring them. Day Time Leaders Day Time Leaders 1 Mon. 3:25 - 4:40 Steph, Jathin 9 Tues. 3:25 - 4:40 Hiatt, Anna V. 2 Mon. 4:50 - 6:05 Sam, Sue 10 Tues. 4:50 - 6:05 EJ, Anna V. 3 Mon. 4:50 - 6:05 Anna K., Hiatt 11 Tues. 4:50 - 6:05 Alexa, Mike 4 Mon. 4:50 - 6:05 Liping, Sameer 12 Tues. 4:50 - 6:05 Zach, Shweta 5 Mon. 6:15 - 7:30 Sue, Anna K. 13 Tues. 6:15 - 7:30 Alexa, Mike 6 Tues. 2:00 - 3:15 Justin, Gurshawn 14 Wed. 3:25 - 4:40 Steph, Jathin 7 Tues. 3:25 - 4:40 Sameer, Gurshawn 15 Wed. 3:25 - 4:40 EJ, Shweta 8 Tues. 3:25 - 4:40 Zach, Justin 16 Wed. 4:50 - 6:05 Sam, Liping PLEASE CIRCLE YOUR WORKSHOP SECTION NUMBER ON THE TABLE BELOW
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2 1. (30 points) Consider the following reaction between N 2 (g) and O 2 (g) : ½N 2 (g) + O 2 (g) NO 2 (g) H f ° kJ mol -1 ) S ° (J K -1 mol -1 ) O 2 (g) 0. 205 N 2 (g) 0. 192 NO 2 (g) 33.10 240 From the data shown above, calculate the following quantities at 1000 ° C. You may assume that H f ° and S ° are independent of temperature. a) H ° = 33.10 kJ/mol b) S ° = 240 – 205 – ½ (192) = -61 J/K-mol c) G ° = H ° - T S ° = 33.1 – (1273)(-61)(10 -3 ) = +110.7 kJ/mol d) K eq : ln K eq = - G ° /RT = -(110,700)/[(8.31)(1273)] = -10.46 K eq = e -10.46 = 2.87 × 10 -5 e) Calculate the enthalpic factor contributing to K eq at 1000 ° C. Provide a qualitative explanation for the value of this factor in terms of the properties of reactants and products. Enthalpic factor = e
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CHM 132 - Exam 2 - 2006 Key - Chemistry 132 Hour...

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