Consider this hypothetical reaction: A + 2B + C --- AB2C
Experimental evidence shows that the reaction occurs in three elementary steps?

Question

Step 1:	A + B ---> AB	      (slow)
Step 2:	AB + B ---> AB2	(fast)
Step 3:	AB2 + C <---> AB2C	(fast, reversible)

(See graphic: http://i270.photobucket.com/albums/jj83/metro_cub/RateMech_zps4d0e5f9d.jpg)

Which rate law is consistent with this mechanism?

(A) rate = k[A][B]

(B) rate = k[AB][B]

(C) rate = k[AB][AB2]

(D) rate = k[A][B]^2[C]

Please explain how you got the answer.  Thank you :-)

ChemTeam · Accepted Answer

The answer is A.

The slow step is the gateway, so to speak, to the rest of the reaction mechanism. As step 1 slowly feeds AB to step 2, the AB immediately reacts with more B. But nothing happening after step 1 can cause it to speed up. So, AB2 and C can never appear in the rate-determining step.

This search:

https://www.google.com/search?q=rate+law+from+reaction+mechanisms&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a&channel=sb

Look at the discussion in the first two hits. They both have an example where the slow step occurs first. it is always the RDS because it controls the flow of the reaction intermediate (AB, in our case) to the next step of the reaction.

HTH.

Jan · Answer

Not sure, but the slow step is rate-determining

I guess (A). Not (B) nor (C).
Maybe (D).

Consider this hypothetical reaction: A + 2B + C ---> AB2C Experimental evidence shows that the reaction occurs in three elementary steps?

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