Intermediates in the rate equation (EK 1001 number 270)

[September24]

---

Members do not see this ad.

Step 1:NO(g)+Br2(g)--k1-->NOBr2(g) (fast)
Step 2:NOBr2(g) + NO (g)--K2-->2NOBr(g) (slow)


270: Which of the following expressions gives the rate of the reaction?

Answer: K2[NObr2][NO]

I know that the rate law is determined by the slow step. However, i thought that the rate law could not include an intermediate. I assumed that [NOBr2] is to be written in terms of [NO] and [Br] or something.

 

[AKMCAT]

Since the first step is the fast step preceding the slow step it should not include the intermediate.
I think the rate of the reaction should be. Is there any explanation provided?
k1*k2 /k-1 [NO]^2 [Br]

 

[deleted380953]

(Editing my original reply) I think that their answer corresponds to the rate of the overall reaction, but not the actual rate law. From my understanding, the rate can always be expressed by the slow step of a reaction, but the rate law is dependent on the mechanism itself and must be determined experimentally. As AKMCAT said, the rate law cannot include intermediates. However it can include either products or reactants. For the link below, I was able to find the exact reaction in your problem on the 3rd slide. It seems like the mechanism itself is whats tested, and if a theoretical rate law can be formed that's on the same order as the experimental one, then it's consistent.

http://g.web.umkc.edu/gounevt/Weblec212Silb/L15(16.6-16.7).pdf

Someone please correct me if I'm wrong. I'd like to know this stuff before Saturday...

 

[September24]

Yeah its quite confusing. I triple checked and the answer they gave is in fact B:k2[NOBr2][NO]
Explanation: The rate of any reaction proceeds at the rate of the slowing step. B gives the rate of the slowest step, Step 2.

....Great. What a simplified explanation. /sarcasm