Exam 1 (Green) - 2010

Exam 1 (Green) - 2010 - CHEM 188 Spring, 2010 Hour Exam 1...

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CHEM 188 – Spring, 2010 Hour Exam 1 (Green) February 11, 2010 Instructions: Your scantron answer sheet must show your NAME , 7-DIGIT KU ID NUMBER , and LAB SECTION . (Begin these entries at the LEFT end of the space provided.) In answering the questions, be careful to fill in the corresponding circles on the answer sheet according to the number of the question on the exam. USE A SOFT (No. 2) PENCIL .  Useful information: Gas constant, R = 8.314 J/K mol = 0.08206 L atm/K mol Integrated Rate Laws: kt [A] 1 [A] 1 : order Second e [A] [A] : order - First kt - [A] [A] : order - Zero 0 kt 0 0 + = - = = - Arrhenius equation: /RT -E a Ae k =               - = 2 1 2 1 a 2 1 T T T T R E k k ln Relation of K P to K C : K P = K C (RT) n Quadratic formula: a ac b b x 2 4 2 - ± - = 1
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1. For the reaction 4NO 2 (g) + O 2 (g) 2N 2 O 5 (g) the rate of loss of molecular oxygen, - d[O 2 ]/dt is 0.0090 M/s at a particular time during the reaction. What is the rate of loss of NO 2 , - d[NO 2 ]/dt ? A. 0.018 M/s B. 0.036 M/s C. 0.072 M/s D. 0.144 M/s E. 0.288 M/s 2. For the overall chemical equation, 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O which one of the following can you rightly assume? A. The reaction is first-order overall. B. The reaction third-order overall. C. The rate law is, rate = k[NO][H 2 ] D. The rate law is, rate = k[NO] 2 [H 2 ] 2 E. The rate law cannot be determined from the information given. 3. The reaction of nitric oxide with hydrogen 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O is second-order in [NO] and first-order in [H 2 ]. At 1280 o C, the rate constant is 25 M - 2 s - 1 . Calculate the rate of the reaction at this temperature if [NO] = 0.20 M and [H 2 ] = 0.20 M . A. 0.050 M/s B. 0.10 M/s C. 0.20 M/s D. 0.30 M/s E. 0.40 M/s 4. Consider the reaction given below. A + B products The following data are obtained at a certain temperature.
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Exam 1 (Green) - 2010 - CHEM 188 Spring, 2010 Hour Exam 1...

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