2006 - Chem 2 Key#1 R = 0.082 L-atm/mol-K = 8.31 J/mol-K...

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Chem 2 Exam #2 March 3, 2006 Key #1 R = 0.082 L-atm/mol-K = 8.31 J/mol-K 1. The following reaction was observed for carbon monoxide and hydrogen: 2 N 2(g) + O 2(g) º 2 N 2 O (g) The concentration of N 2 was measured as a function of time. The rate of the reaction is equal to: a) b) c) rate N t =− Δ Δ [] 2 rate N t = Δ Δ 2 rate N t = 2 2 Δ Δ d) e) rate N t = Δ Δ 2 2 rate N t Δ Δ 2 2 2. Using the data below, estimate the rate of the reaction at 3.0 s for the reaction in Question #1. time, s [N 2 ], atm 0 0.500 1.0 0.496 2.0 0.492 3.0 0.488 4.0 0.484 a) 0.488 atm/s b) -0.012 atm/s c) 0.081 atm/s d) 0.0020 atm/s e) 0.0040 atm/s

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3. Ferricyanide ([Fe(CN) 6 ] 3- ) reacts with hydroxylamine in the following reaction: 2 [Fe(CN) 6 ] 3- + 2 NH 2 OH 6 2 [Fe(CN) 6 ] 4- + N 2 + 2 H 2 O + 2 H + Experiment # [[Fe(CN) 6 ] 3- ], M [NH 2 OH], M [H + ], M initial rate, M /s 1 2.8 x 10 -4 8.0 x 10 -3 1.0 x 10 -5 2.2 x 10 -4 2 4.0 x 10 -4 8.0 x 10 -3 1.0 x 10 -5 3.1 x 10 -4 3 4.0 x 10 -4 8.0 x 10 -3 1.0 x 10 -4 3.1 x 10 -5 The rate law can be written as: rate = k [[Fe(CN) 6 ] 3- ] m [NH 2 OH] n [H + ] p What is the value of m? a) -1 b) 0 c) +1 d) 2 e) +1/2 4. Using the data in Question #3, what is the value of p? a) -1 b) 0 c) +1 d) 2 e) +1/2
5. For the following reaction: NO 2(g) + CO (g) 6 NO (g) + CO 2(g) The rate law was found to be: rate = k [NO 2 ] 2 The initial rate was found to be 6.0 x 10 -3 atm/s when the pressure of NO 2 was 2.0 atm and the pressure of CO was 0.20 atm. What is the value of k? a) 3.0 x 10 -3 atm -1 s -1 b) 1.5 x 10 -3 atm -1 s -1 c) 7.5 x 10 -3 atm -1 s -1 d) 2.4 x 10 -2 atm -1 s -1 e) 4.8 x 10 -3 atm -1 s -1 6. For the reaction in Question #5, what would be the initial rate if the pressure of NO 2 were increased to 5.0 atm? a) 6.0 x 10 -3 atm/s b) 2.4 x 10 -3 atm/s c) 1.5 x 10 -2 atm/s d) 3.8 x 10 -2 atm/s e) 1.2 x 10 -2 atm/s 7. For the reaction in Question #5, what would be the initial rate if the pressure of NO 2 remained at 2.0 atm and the pressure of CO increased to 0.80 atm? a) 6.0 x 10 -3 atm/s b) 2.4 x 10 -2 atm/s c) 4.8 x 10 -3 atm/s d) 1.5 x 10 -3 atm/s e) 7.5 x 10 -3 atm/s

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8. The reaction in Question #5 was studied at a higher temperature and the value of the rate constant was found to be 2.0 atm -1 s -1 at 600 K. The rate constant at 300 K can be estimated to be: a) about 2.0 atm -1 s -1 b) about 1.0 atm -1 s -1 c) at least 10 times greater than 2.0 atm -1 s -1 d) at least 10 times smaller than 1.0 atm -1 s -1 e) cannot be estimated from the data 9. The following reaction was studied at different temperatures: 2 N 2 O (g) 6 2 N 2(g) + O 2(g) A plot was made with ln k on the y-axis and 1/T on the x-axis. The slope of this graph was found to be -2.89 x 10 4 K. What is the activation energy for this reaction? a) -2.37 kJ b) +2.37 kJ c) 3.48 kJ d) 352 kJ e) 240 kJ 10. The following reaction has been observed: O 2 + 2 NO 2 - 6 2 NO 3 - One mechanism for this reaction is: Step 1: O 2 + NO 2 - 6 O 3 + NO - Step 2: O 2 + NO - 6 NO 3 - Step 3: O 3 + NO 2 - 6 O 2 + NO 3 - We will assume that Step 3 is fast. The empirical rate law is: rate = k [O 2 ] 2 [NO 2 - ] If this mechanism is true, the slow step in the reaction is: a) Step 1 b) Step 2 c) it is not possible from this information to determine the slow step
11. The ) H for the following reaction is 64.7 kJ and the energy of activation (E

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This note was uploaded on 04/07/2008 for the course CHEM 002 taught by Professor Ryan during the Spring '08 term at Marquette.

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2006 - Chem 2 Key#1 R = 0.082 L-atm/mol-K = 8.31 J/mol-K...

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