Chemistry Chap. 6 Practice Passage Q.4

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ja270
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Joined: Tue Oct 09, 2018 9:07 pm

Chemistry Chap. 6 Practice Passage Q.4

Post by ja270 » Sun Oct 21, 2018 6:24 pm

Hello,

I was wondering if I could get some help with question four of the practice passage. In particular I'm confused about why/how we combine the fast and slow steps for I. Is there some background/outside information about rate laws for reversible reactions that I don't remember?

Thanks!
NS_Tutor_Will
Posts: 75
Joined: Fri May 25, 2018 9:15 am

Re: Chemistry Chap. 6 Practice Passage Q.4

Post by NS_Tutor_Will » Thu Oct 25, 2018 2:00 pm

For I, a product from first reaction (fast) becomes a reactant in the second reaction (slow). So we solve the rate law expression for that value (N2O2) and then plug that value into the rate law expression for the second step.

We solve the first rate law expression to give:
[N2O2] = k1/k1 [NO]2

Then we plug that N202 here:
k2 [N2O2][O2]

To yield:
k2 k1/k1 [NO]2[O2]

Then, we can simplify that to
k [NO]2[O2] = rate

Which is consistent with the data in Table 1.

I hope this helps!
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