15. 89


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Meghna2A
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15. 89

Postby Meghna2A » Tue Mar 13, 2018 10:17 pm

How do we know that step 2 is slow and step 3 while 1 is in equilibrium, before we substitute with step 1?
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Farah Ahmad 2A
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Re: 15. 89

Postby Farah Ahmad 2A » Tue Mar 13, 2018 10:23 pm

We know Step 2 is the slow step instead of Step 3 because N2O2 is an intermediate, which is why we would have to use the pre-equilibrium method and substitute in [NO]^2 for [N2O2]. If step 3 were the slow state N2O would have to be replaced by H2 and N2O2, which does not match with the rate law given.
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