exam_3_review-_answers

exam_3_review-_answers - EXAM 3 REVIEW- PRACTICE QUESTION...

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EXAM 3 REVIEW- PRACTICE QUESTION ANSWERS LBS 172 REACTION MECHANISMS 1.) Intermediates: N 2 O 2 (g), N 2 O(g), Steps 1, 2, and 3 are each bimolecular, Rate step 3 = k[N 2 O][H 2 ]. The steps add up to the overall reaction. 2.) Rate mech A = k step 1 [NO 2 ], Rate mech B = k step 1 [NO 2 ] 2 Since the rate equations for each mechanism depends differently on the concentration of NO 2 , one could examine [NO 2 ] over a time interval to determine if the reaction is 1 st or 2 nd order (or other), using rate laws with respect to NO 2 to eliminate the possibility of one or both of the mechanisms. 3.) Rate = k 2 [NO 2 Cl][Cl]. Cl is an immeasurable intermediate, so [Cl] must be rewritten as [Cl] = k 1 [NO 2 Cl] / (k -1 [NO 2 ]). Finally, Rate = k 2 [NO 2 Cl] k 1 [NO 2 Cl] / (k -1 [NO 2 ]) 4.) The diagram represents a two step exothermic reaction, as there are two peaks and the products have less energy than the reactants. If there is enough energy to reach the second peak (slow step), then there is already plenty of energy to go forward and backward across the first peak (fast step). Sufficient energy for a forward and backward reaction of step one allows for equilibrium. 5.) Heating the sample increases proportion of molecules with enough energy to react, increasing the rate of the reaction. Since the concentrations of reactants are not directly affected by the temperature change, k must increase to mathematically increase the rate
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exam_3_review-_answers - EXAM 3 REVIEW- PRACTICE QUESTION...

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