[LaChimie]

Hello all!  I was hoping you might be able to help me out with this problem about developing a rate equation for this reaction.  This problem is unlike the examples we went over and class, and I'm a bit confused how to do it.  

Consider the aqueous reaction (consider using the abbreviation Ph = C6H5 ) :
H+ + HNO2  +  C6H5NH2  -->  C6H5N2+  +  2H2O    
  with mechanism:
(1)    H+ + HNO2    <-->    H2NO2+        (fast equil)   

(2)     H2NO2+  + Br-   -->    ONBr + H2O      (slow)

(3)     ONBr + C6H5NH2    -->    C6H5N2+  +  H2O + Br-     (fast)    

  Questions:
(a)  What specie/s do you expect to be the catalyst/s in the mechanism?   
(b)  What specie/s is/are the intermediate/s in the mechanism?
(c)  Use quasi-equilibrium analysis to derive a rate law from the mechanism
that could be matched up with experiment, i.e. eliminate any intermediates.  

I'm inclined to say that Br- is the catalyst in this mechanism, but I can't seem to derive a rate law that matches this assumption (i.e. the reaction rate is zeroth order in Br-).  

[renge ishyo]

If Br- is a catalyst in the reaction as you suggest, then it would not appear in the rate law, and you would get a rate law that looks something like this:

rate = k[H₂NO₂⁺]

Now you are left with the task of replacing the intermediate in the rate law; the process should be similar to some of those examples you saw in class