Lecture 9

Lecture 9 - 1/25/10 Consider the reaction 2 NO + O2 (g) ! 2...

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1/25/10 1 Consider the reaction 2 NO + O 2 (g) ! 2 NO 2 (g) Observed rate law : Rate = k[NO] 2 [O 2 ] The proposed mechanism : Step 1) NO + O 2 NO 3 (g) Step 2) NO (g) + NO 3 (g) ! 2 NO 2 (g) (Slow) 93 Which of the following is the reaction profile for this reaction? B) A) C) D) RDS (Fast equilibrium) ! k[NO][NO 3 ] NO 3 intermediate: can’t be in rate law First. Write rate laws for fast step (both directions) and slow step Rate 1 (forward) = k 1 [NO][O 2 ] 94 Step 2) NO (g) + NO 3 (g) ! 2 NO 2 (g) Step 1) NO + O 2 NO 3 (g) (Fast equilibrium) ! Rate 1 (reverse) = k 2 [NO 3 ] Rate 2 = k 3 [NO][NO 3 ] But cannot be overall rate law NO 3 can’t be in rate law Overall rate = Second. Express intermediate in terms of reactants by making k 1 [NO][O 2 ] = k 2 [NO 3 ] [NO 3 ] = k 1 [NO][O 2 ] k 2 95 rate 1 (forward) = rate 1 (reverse) = k 3 [NO] k 1 [NO][O 2 ] k 2 where k 1 k 3 /k 2 = overall k Third. Substitute the expression for intermediate into rate law for rate-determining step k 2 = k 1 k 3 [NO] 2 [O 2 ] Rate 2 = k 3 [NO][NO 3 ] Rate = k[NO] 2 [O 2 ] True if Step 1 reaches equilibrium 7. Catalysis : speeding up a reaction (Section 16.8) How can we increase the rate of the reaction?
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Lecture 9 - 1/25/10 Consider the reaction 2 NO + O2 (g) ! 2...

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