ch15 #51

$K = \frac{k_{forward}}{k_{reverse}}$

Claire_Zhou_1A
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ch15 #51

For #51, the question is write the rate law for the formation of NOBr
step 1 NO + Br2 → NOBr2 (slow)
step 2 NOBr2 + NO → NOBr + NOBr (fast)
the answer says because the first step is the rate-determining step so the rate of formation of NOBr=k[NO][Br2].
I thought according to step 1 rate of formation of NOBr2 = k1[NO][Br2] and then substitute this expression into step 2, rate of formation of NOBr = 2k2[NOBr2][NO]. I got the answer rate=2k1k2[NO]²[Br]. Could someone explain to me why I am wrong? And in general, what should we do if a fast step is after the slow step?

Nerissa_Low_2F
Posts: 23
Joined: Fri Jul 15, 2016 3:00 am

Re: ch15 #51

I think the rate law is only governed by the slow step, so the overall rate law would be that of the first step.

Cecile Riviere-Cazaux 2A
Posts: 11
Joined: Fri Jul 15, 2016 3:00 am

Re: ch15 #51

Because the first step is the slow step, the entire reaction mechanism is going to be controlled by this. Your method is used when you have a fast step and then a slow step because there is a bottleneck effect resulting in the buildup of intermediate products which is why the first step is approximately in equilibrium and why you can use that approach. The slowest step (step 1) determine the rate of the overall reaction, which is why the answer to 51 is rate= k[NO][Br2].

Samuel_Vydro_1I
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

Re: ch15 #51

Yes, as the questions on the Quizzes, the slow step governs the entirety of the reaction. When you see a reaction that is slow then fast, ONLY look at the slow step to determine rate law. Nice general rule