rate of reaction

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chiddler

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Answer: A

My question: Why isn't B the answer?

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Only did this quickly in my head, but if you add them all together the only things that don't cancel out are A2 and 2BC (which ends up as BC^2).
 
Where did you get this question from?

From the looks of it, the answer they want takes into account all of the reactants before the slow step (B doesn't count, because it's also a product).

Edit: after some searching around: you're right. Intermediates don't go into the rate law. Rather, you have to express their concentrations in terms of the reactants that formed them. http://www.science.uwaterloo.ca/~cchieh/cact/c123/ratemech.html

So k2[BC] needs to have that 'translated'; and its concentration is k1[A2][BC]. Thus the rate law k'[A2][BC][BC]
 
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The formation of B2C requires the formation of B.
The formation of B has a rate of [A2][BC]

It's just substitution.
 
Why can't you use Hess's law to get the net reaction?
 
No. You can't have an intermediate occur in your rate law. A2C is an intermediate. The right answer is A, which is in terms of only reactants and takes into account the double dependence on the concentration of BC.

i'm talking about question in the link.
 
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