[Ritted]

Hey Guys,

I was wondering if anyone knew how to solve the following question.  (I believe that the answer is C; however, my Chemistry professor believes that the answer is A) 
      Step 1: N2H2O2  N2HO2- + H+   (fast equilibrium)
      Step 2: N2HO2-  N2O + OH-   (slow)
      Step 3: H+ + OH-  H2O      (fast)
Nitramide, N2H2O2, decomposes slowly in aqueous solution.  This decomposition is believed to occur according to the reaction mechanism above.  The rate law for the decomposition of nitramide that is consistent with this mechanism is given by which of the following?
(A) Rate= k[N2H2O2]
(B) Rate= k[N2H2O2][H+]
(C) Rate= k[N2H2O2]/[H+]
(D) Rate= k[N2H2O2]/[N2HO2-]
(E) Rate= k[N2H2O2][OH-]

Thanks!  I really appreciate it!

-Ritted

[Ritted]

*Ignore me, I am impatient*

[cliverlong]

Maybe this will help

http://www.chemguide.co.uk/physical/basicrates/ordermech.html#top

I agree with your professor's answer. Why?

Clive

[Ritted]

Well, I googled around for the question and found this: http://dipowell1.home.mindspring.com/apmultiplechoice/topic13.doc

This document suggests that I'm correct, but I'm not sure how to justify my answer properly.

[Astrokel]

The slowest step determines the rate law and K₊/K_- = constant, try eliminating the intermediate and you will get C.

[cliverlong]

Well, I googled around for the question and found this: http://dipowell1.home.mindspring.com/apmultiplechoice/topic13.doc

This document suggests that I'm correct, but I'm not sure how to justify my answer properly.

As you wrote, this sheet gives you an answer to your question but does not justify it - so why do you think it is correct?

I will give an argument for why I think A is correct. I may be wrong.


The rate equation only includes reactants that you introduce it the beginning of the reaction. You cannot measure intermediates so they can't appear in the rate equation (eliminate answers B and E)
The rate equation is composed of the product of concentration of reactants raised to certain powers - there is no division involved as in equilibrium equation (eliminate answers C and D)
The rate is determined by the slowest step in the reaction. If an internediate reaction is slow then all the reactions before it are also slowed up. As you can only measure the concentrations of the initial reactant, it appears that in this reaction is the concentration of the only reactant that is the limiting factor in the rate - hence A is the answer.

Clive

[Astrokel]

http://www.google.com.sg/search?hl=en&q=rate+law+with+intermediate&meta=&aq=f&oq=

[cliverlong]

http://www.google.com.sg/search?hl=en&q=rate+law+with+intermediate&meta=&aq=f&oq=

Yes, that all makes sense.

In none of those examples does the rate equation have a division.

In my answer above I argued C has a division therefore answer C cannot be a rate equation, let alone the correct rate equation.

Clive

[Ritted]

The problem is that A cannot be the answer because, if it were, then the first step would have to be the rate-determining step.  However, we know that the first step is not the rate determining step because it's an extremely fast step. 

[cliverlong]

The problem is that A cannot be the answer because, if it were, then the first step would have to be the rate-determining step.  However, we know that the first step is not the rate determining step because it's an extremely fast step.  

If you read carefully the link I have given and the example given by Astrokel you will find that is not a valid argument. You are confusing rate determining step with rate equation.

I emphasise again that a slow step "down the line" will cause the slow rate to "back up" to the original reactants. Think of a lot of people  in a department store or shop and there is only one small double door that will alow people in and out. The fire alarm goes off and the people in the store can run to the doors, so they can move fast. However, at the door thery are restricted to only two people moving at the same time. If we measure the rate of movement in the store rather than at the door you will find there is a slow rate of movement even though there is loads of space and people can run. (and a few people from the outside will still be trying to get into the store because there is a sale on) In writing a rate equation the only concentrations that appear are the reactants, (the first equation) not intermediates (subsequent equations). Think about the reaction of 3 hydrogen molecules with one nitrogen molecule and find a rate equation. Do you think the reaction proceeds in one step where 4 molecules collide? Does the rate equation identify any intermediates or only hydrogen and nitrogen?

Clive

[Ritted]

Your argument seems valid; however, after further investigation, I have found that this question is from a released AP Chemistry exam (an international high school chemistry exam), and the answer the test writers gave was C as well. 

[Astrokel]

The slowest step will always be the RDS and the division does not matter as it is just an expression obtain to get rid of the intermediate. I would not consider proton as intermediate as it is used it the last step. Furthermore the RDS is not in an equilibrium thus making the rate law even less direct. Correct my arguement if i am wrong.

[cliverlong]

The slowest step will always be the RDS

I agree, and I have written so above - giving a silly fire alarm in a store analogy.

and the division does not matter as it is just an expression obtain to get rid of the intermediate.

Yes it does matter. You can't use an algebraic manoeuvre if it leads to an incorrect answer

I would not consider proton as intermediate as it is used it the last step.

Yes it is an intermediate. The fact it is used in the last step is irrelevant.
The confusion, I believe arises, because the possible reaction steps have all been laid out at the start.

If we look at the overall equation of the decomposition of nitranamide

N₂H₂O₂  N₂O + H₂0

H⁺ is not visible in the overall equation, and it definitely isn't an initial reactant, so it can't be part of the rate equation. This behaviour is seen in all the examples linked to in the above posts.

Furthermore the RDS is not in an equilibrium thus making the rate law even less direct.

Sorry, I don't see your point

Correct my argument if i am wrong.

I believe I have

I may be wrong but I think I have laid my argument out clearly in the posts above. I would be very grateful if anyone can expose a flaw in my argument by pointing out which bit is wrong and providing a reference as to which part of my explanation is wrong. Currently I'm confident in my argument for the reasons given in my several posts above.

Clive