U6_Exam_S10_KEY - KEY- EXAM 2- CHEM XXI-152 SPRING 2010 Dr....

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KEY- EXAM 2- CHEM XXI-152 SPRING 2010 Dr. Pollard 1. Nitrogen monoxide (NO) is a pollutant commonly formed by the reaction of oxygen (O 2 ) and nitrogen (N 2 ) inside our cars’ combustion engines: N 2 (g) + O 2 (g)  2 NO(g) The NO(g) produced reacts with O 2 (g) to produce NO 2 (g) following a two-step mechanim: NO(g) + O 2 (g) OONO(g) Fast equilibrium NO(g) + OONO(g) 2 NO 2 (g) Slow step a) (5) Write the overall chemical equation for this process: 2 NO(g) + O 2 (g) 2NO 2 (g) All or nothing b) (5) Identify all the intermediates in this mechanism. OONO(g) All or nothing c) (15) Use the mechanisms to derive the rate law for this. Clearly show the work done to derive the answer. Sketch a reaction diagram of potential energy vs. reaction progress. Indicated the location of any intermediates and label the activation energies. The reaction is endothermic. The reaction is actually exothermic due to A-A to A-B bonds, so the endothermic is an error. You can draw it as exothermic and still get credit. Rate = kK c [NO] 2 [O 2 ] OR JUST Rate = kNO] 2 [O 2 ] (7 points, 7, 3 or 0. 3 for having 1 error). No work, no credit. 8 points. 1 pt for labeling intermediate, 2 pts for labeling each Ea, 5 points for shape of curve. d) (5) By what factor will the rate of this process decrease if a catalytic converter reduced the concentration of NO(g) in half in the combustion engine? Show the work and your thinking. . By a factor of 4 since it is second order. (3 for factor, 2 for justification) OONO Ea1 Ea1 PE Reaction Progress
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2. The pollutant NO 2 (g) may decompose back into NO(g) and O 2 (g) according with the following reaction: NO 2 (g) 1/2 O 2 (g) + NO(g) a) (10) The table shows how the concentration of NO 2 (g) changes as a function of time inside a
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U6_Exam_S10_KEY - KEY- EXAM 2- CHEM XXI-152 SPRING 2010 Dr....

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