rate of a reaction

[rocky90]

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If we have a reaction such as:
NO2+N02-----> N2O4 (fast)
N2O4+H2---->N2O4+H20 (slow)

Wouldnt the rate of the reaction be [NO2]2 [H2] since N2O4 is an intermediate. I was doing a very similar question in Destroyer or topscore (dont remember exactly) and the answer said [N2O4] [H2].
Thanks

 

[buckey]

the rate of the reaction is found by using the substrates of the slow step. the slow step is the reaction with N2O4 and H2 so those are used for the rate expression

 

[rocky90]

But the rate of a reaction is never dependent on the intermediate. Hence this should not be correct. At least thats what I have learnt.
Thanks

 

[rocky90]

anyone please urgently. I dont want this to show on the test without me being sure what to answer and this is pretty simple.

 

[DentistDMD]

But the rate of a reaction is never dependent on the intermediate.

A counterexample would be a carbocation, which is an intermediate that acts as a rate determining step, formed during the SN1 reaction. The rate determining step is simply a mechanism of the slowest step as the above poster had mentioned. And the rate equation would be the k * reactants of the slowest step.

 

[FROGGBUSTER]

But the rate of a reaction is never dependent on the intermediate. Hence this should not be correct. At least thats what I have learnt.
Thanks

Not true, the rate of many reactions are dependent on intermediates and intermediate steps.

 

[rocky90]

So the answer to it would be that rate=k[N2O4] [H2] right rather than [NO2]2 [H2]. In my class I remember our professor told us that and this example I posted is from his notes. Just want to confirm as I am sure this can show up in the DAT test.

 

[rocky90]

I am actually right I think and the answer in topscore/destroyer is wrong. For a rate determining step intermediates are never involved. Here is the link.
Thanks.

http://bouman.chem.georgetown.edu/S02/lect5/lect5.htm
http://au.answers.yahoo.com/question/index?qid=20110704051050AAoiv9t
http://answers.yahoo.com/question/index?qid=20080610204445AA3TNBi

 

[FROGGBUSTER]

I am actually right I think and the answer in topscore/destroyer is wrong. For a rate determining step intermediates are never involved. Here is the link.
Thanks.

http://bouman.chem.georgetown.edu/S02/lect5/lect5.htm
http://au.answers.yahoo.com/question/index?qid=20110704051050AAoiv9t

From what I recall, it is that rate laws are preferentially not expressed in terms of intermediates, but that does not mean intermediates do not factor into the rate of multi-step reactions.

NO2+N02-----> N2O4 (fast)
N2O4+H2---->N2O4+H20 (slow)

Rate = k * [N2O4] * [H2].

You now have to substitute the NO2 concentrations for [N2O4].

Because the fast step precedes the slow step, the fast step is in equilibrium. This means kforward * [NO2]^2 = kreverse * [N2O4]

This means [N2O4] = kforward/kreverse * [NO2]^2 = K * [NO2]^2, where K = equilibrium constant and k = rate constants.

This means Rate = k*kforward/kreverse * [NO2]^2 * [H2].


BTW, this will never show up on the DAT.

 

[rocky90]

You are exactly right. I just dont like the fact that people who make tests for topscore/ destroyer post wrong explanation for wrong answers.
Moreover say if you had both the answers in your choose what would you pick? I picked the answer I knew was right and still got the question wrong. I hope and pray this doesnt happen in DAT like you said.