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Chemistry 8 Online
OpenStudy (anonymous):

can someone help with this question

OpenStudy (anonymous):

OpenStudy (petiteme):

I don't know. I forgot how to do this. Sorry.

OpenStudy (anonymous):

do u know who does. bc i dont know how to do it

OpenStudy (anonymous):

do it by the electronic formulase , means do it by forgot .. sorry

OpenStudy (wolfe8):

I think you somehow rearrange equations 1, 2, and 3 to get the overall reaction as the question. And note what you do to the equations so you will do the same with their k's... Sorry that's the best I got. Petite asked me last night and I said the same thing she did.

OpenStudy (aaronq):

Because the second reaction is the rate-determining step, you only have to take into account the first 2 reactions. Product from the first reaction gets accumulated at the second step (because it's slow), as a result, some gets turned back into the first reactant, O2NNH2. The rate of the overall reaction is the sum of both rate constants for the forward reactions (k1 and k2) and over the constant for the backward reaction (k-1), so k= (k1)+(k2)/(k-1).

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