1ea 3 ccircleanycatalysts 1uncatalyzed

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Unformatted text preview: 1 +1 correct answer, +1 correct units 6 +2 all or nothing 7. Below are the mechanisms for an uncatalyzed and a catalyzed reaction. k1 k1 Step 1: NO(g) + O3(g) NO2 + O2(g) (slow) Step 1: O3(g) O2(g) + O (g) (fast) k­1 k2 k2 Step 2: O(g) + O3(g) 2 O2(g) (slow) Step 2: O3(g) O2(g) + O (g) (fast) k3 rate = k2[O][O3] but O is an intermediate! +2 Step 3: NO2(g) + O(g) NO(g) + O2(g) (fast) To substitute: +2 equilibrium rate rate = k1[NO][O3] +2 k1[O3] = k‐1[O2][O] [O] = k1[O3] k‐1[O2] 2 rate = k2k1[O3] +1 overall rate k‐1[O2] 5 2 (a) Write the overall rate law for each mechanism +1 # of hills (b) Put a box around each intermediate. +1 Ea 3 (c) Circle any catalysts. +1 uncatalyzed (d) Draw both mechanisms on the same reaction is higher coordinate diagram, paying attention to the relative scale of the reactions. Assume the reactions are exothermic. reactants +1 exothermic (e) Label the activation energy for the +1 # of hills 6 +1 Ea first step in each mechanism in the diagram. products reaction coordinate 4 A 8. The reaction 2 NO2(g) + F2(g) 2 NO2F(g) has an overall second order rate law, rate = k[NO2][F...
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This note was uploaded on 09/20/2012 for the course CHEM 1252 taught by Professor Jew during the Fall '12 term at UNC Charlotte.

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