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Postby kanikubari » Tue Mar 12, 2019 8:18 pm

Why is mechanism C not correct. How does step 1 having a reverse that is both fast and at equilibrium affect this?

A De Castro 14B 2H
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Re: 15.53

Postby A De Castro 14B 2H » Tue Mar 12, 2019 9:17 pm

Mechanism C wouldn't agree with the proposed rate law for the same reason that mechanism A wouldn't: the rate law would include CO. Having a reverse reaction in step 1 means that the reactants used to produce NO3 can be used in the rate law. Thus, for (c), rate = k[NO2]2[CO]. But, again, this doesn't agree with the proposed rate law, rate = k[NO2]2.

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