Exam 2 - CHEMISTRY 116 Exam 2 Dr T.R Chigwada An optical...

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CHEMISTRY 116 Exam 2 Dr. T.R Chigwada July 19, 2007 An optical scoring machine will grade this examination. The machine is not programmed to accept the correct one of two or more sensed answers and will not sense answers that are lightly marked. Mark your sheet carefully with a No. 2 soft lead pencil and erase any undesired marks COMPLETELY . Avoid making any extraneous marks on the answer sheet other than the information mentioned below. On the answer sheet: 1. Print your name in the space for NAME (last name first, CIRCLE your last name). 2. In the space marked SUBJECT write Chem 116. 3. In the space marked TEST NO. write Exam 2. 4. In the space DATE write Summer 5. In the space marked PERIOD write Chigwada. 6. Check to see that you have 21 examination questions, a periodic table, and a scantron. Hand in only the answer sheet!! Useful Equations and Constants: R = 0.08206 L atm/mol K = 8.314 J/mol K 1 atm = 760 torr = 760 mm Hg ln [A] t = ln[A] 0 – kt [A] t /[A] 0 = e – (kt) [A] t = [A] 0 – kt 1/[A] t = 1/[A] 0 + kt ln k = ln A – E a /RT k = Ae -Ea/RT ln(k 2 /k 1 ) = –E a /R[(1/T 2 ) – (1/T 1 )] Relax and Good Luck !!!!!
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1. In a particular study of the reaction described by the equation, 2 CH 4 O( g ) + 3 O 2 ( g ) 2 CO 2 ( g ) + 4 H 2 O( g ), the rate of consumption of O 2 ( g ) is 0.400 mol L 1 s 1 . What is the r ate of formation of H 2 O ( g ) in the study? A) 0.300 mol L 1 s 1 B) 0.400 mol L 1 s 1 C) 0.533 mol L 1 s 1 D) 0.800 mol L 1 s 1 E) 1.33 mol L 1 s 1 2. The reaction, 2 NO( g ) + O 2 ( g ) 2 NO 2 ( g ), was found to be first order in each of the two reactants and second order overall . The rate law should therefore be written as A) rate = k[NO] 2 B) rate = k([NO][O 2 ] C) rate = k[NO 2 ] 2 [NO] 2 [O 2 ] ½ D) rate = k[NO] 2 [O 2 ] 2 E) rate = k([NO][O 2 ]) 2 3. For the reaction, 3 B + C
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Exam 2 - CHEMISTRY 116 Exam 2 Dr T.R Chigwada An optical...

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