## 15.51? Molecularity?

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

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### 15.51? Molecularity?

For 15.51, I don't understand why the overall reaction ( 2NO+Br2 --> 2NOBr) is considered to be bimolecular according to the solutions manual? Wouldn't it be trimolecular, which makes the rate law= k[NO]^2[Br2] ? I thought you include the coefficients when writing a rate law for a reaction and also determining its molecularity.

Sabrina Dunbar 1I
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### Re: 15.51? Molecularity?

The rate law is determined by the slow step, which in this problem is NO+Br2-->NOBr2. The rate law based on this step in the process is rate=k[NO][Br2] and the molecularity is two because there is one molecule of each NO and Br2 colliding to create the intermediate NOBr2. Hopefully this helps!

Julianna Thrasher 1B
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### Re: 15.51? Molecularity?

I agree with Sabrina. To elaborate, if the slow step is the first step (not the second), then it is the only step that matters for the overall reaction order. The second step (fast) occurs too quickly to modify the rate law and also because it occurs after the slow step. The slow step will always be the rate controlling step, and in this case it occurred as the first step-- compared to many problems we have seen it as the second step, which is harder and makes us consider both steps.

April P 1C
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### Re: 15.51? Molecularity?

The rate law of any reaction is always given by the slow step. Therefore, because the second step is fast, you can ignore the second step.The Rate law for the formation of NOBr will then be Rate=k[NO][Br2] since the slow step says NO +Br2 -> NOBr2