Problem Set 1 Solutions

Problem Set 1 - Chem 102 Klobukowski Winter 2009 Problem Set 1 Solutions Unit 1 1 A key reaction in the upper atmosphere is O3(g O(g 2 O2(g The

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Chem 102 Klobukowski Winter 2009 Problem Set 1 Solutions: Unit 1 1. A key reaction in the upper atmosphere is : O 3 (g) + O (g) 2 O 2 (g) The activation energy for the forward reaction is 19 kJ and the change in enthalpy for the reaction is –392 kJ. Draw a reaction energy diagram (potential energy as a function of reaction progress) for this reaction, postulate the structure of the activated complex, and calculate the activation energy for the reverse reaction. ANSWER: The transition state is formed when O 3 and O collide : _ O=O O O (in reality, this structure is bent because of lone pair) Atoms indicated in bold are from ozone. The atomic oxygen must approach the ozone molecule end-on so that one of the end atoms of ozone can combine with O to form O 2 . There is one bond that is partially made and one that is partially broken (indicated by dotted lines), the double bond remains throughout the transition state. The oxygen atom does not approach the central atom of ozone as this can’t simply lead to the formation of O 2 . 2. The following overall reaction : 2 NO 2 (g) + F 2 (g) 2 NO 2 F (g) Δ H < 0 has this accepted mechanism: Step 1 NO 2 (g) + F 2 (g) NO 2 F (g) + F (g) slow (RDS) Δ H > 0 Step 2 NO 2 (g) + F (g) NO 2 F (g) fast Δ H < 0 Activated complex #1 is the highest energy species, and F is bound to N in NO 2 F. (a) Draw a reaction energy diagram and postulate the structure of the species at both transition states. (b) Predict the overall rate law for this reaction. (c) Determine what species are present initially, at the transition states, at the end of step 1, and at the end of the overall reaction. (d) What is the intermediate? E a = 19 kJ Pot. E E a (reverse)= 392 kJ + 19 kJ = 411kJ Δ H rxn = -392 kJ
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Chem 102 Klobukowski Winter 2009 ANSWER: (a) (* indicates end of step 1) TS 1 TS 2 Pot. En. * Δ H rxn < 0 Reaction progress Transition state complex #1 (TS1): Transition state complex #2 (TS2): (b) Overall rate law for this reaction is determined by step 1, the slowest step. Rate = k [NO
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This note was uploaded on 03/25/2009 for the course CHEM 102 taught by Professor Drk during the Spring '09 term at University of Alberta.

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Problem Set 1 - Chem 102 Klobukowski Winter 2009 Problem Set 1 Solutions Unit 1 1 A key reaction in the upper atmosphere is O3(g O(g 2 O2(g The

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