F03_E4A - Chemistry 111 Page 1 out of 8 Exam IV /Fall 2003...

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Chemistry 111 Page 1 out of 8 Exam IV /Fall 2003 Answer Key 1. Look at the following mechanism and answer the questions below. A 2 B + X A 2 BX k 1 k 2 k 3 A 2 BX 2 A + B + X a) What would you call the substance X? A catalyst . (02 pts) b) What would you call the substance A 2 BX? An intermediate . (02 pts) c) What is the overall reaction? A 2 B 2 A + B (02 pts) 2. Consider the following reaction, which takes place at 25°C. 2 NO 2 + F 2 2 NO 2 F Experimentally, it is observed that the rate equation is second order, first order with respect to each reagent: rate = k [NO 2 ] [F 2 ]. Propose a two step mechanism for this reaction. Indicate which is the slow step of the mechanism. (05 pts) The important thing about this question is to realize that the experimental rate equation tells us the number of molecules that collide simultaneously in the slow step of the mechanism. Therefore, no matter what else we do, we should immediately realize that in the slow step we MUST HAVE one molecule of NO 2 reacting with one molecule of F 2 . Therefore, a plausible mechanism would be: Step #1: NO 2 + F 2 NO 2 F + F Slow Step #2: NO 2 + F NO 2 F Fast 3. In the previous exam, Spiderman was in trouble and some of you suggested slowing down the reaction by lowering the temperature of the room. Assuming that the activation energy of the reaction was 21.0 kJ/mol and that the initial temperature of the room was 27°C, what would have had to be final temperature in order to double the time he had to get out of the room? (10 pts) From the PUI, we know that…
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Chemistry 111 Page 2 out of 8 Exam IV /Fall 2003 Answer Key ln k 2 k 1 = Ea R × 1 T 2 1 T 1 In order to double the time, we have to slow down the reaction (so it will take twice as long). This means that we want the ratios of the k’s to be one half… Therefore, ln 0.500 () = 21,000 J/mol 8.314 J/mol K × 1 T 2 1 300 K T 2 = 277 K or approx. 4°C . 4. A certain student measured experimentally the heat capacity (Cp) for a certain gas and obtained 43.2 J/mol K. However, he needed the value of Cv. His lab partner told him not to worry because since the gas is treated as ideal, then the Cv would have been 51.5 J/mol K. Is this true or false ? Circle your answer. (02 pts) False ! The value for Cp for an ideal gas is always larger than the value for Cv. Therefore, the Cv would have been 34.9 J/K mol instead. 5. In another experiment, a student was given an unknown gas and was told that it was either N 2 , O 2 , or Xe gas. He measured the heat capacity of the unknown gas (Cp) and determined it to be 20.8 J/mol K. Can he identify the gas based on this only? Explain!
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F03_E4A - Chemistry 111 Page 1 out of 8 Exam IV /Fall 2003...

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