## 15.53 (c)

$K = \frac{k_{forward}}{k_{reverse}}$

Hannah Guo 3D
Posts: 56
Joined: Fri Sep 29, 2017 7:06 am

### 15.53 (c)

Three mechanisms for the reaction NO2(g) CO(g) + CO2(g) --> NO(g) have been proposed:
(c)Step1 NO2 + NO2 ⇌ NO + NO3 and its reverse (both fast, equilibrium)
Step2 NO3 + CO --> NO2 + CO2 (slow)
Which mechanism agrees with the following rate law: Rate = k[NO2]^2? Explain your reasoning.

The answer says this is false. Rate=k[NO3][CO], but [CO] cannot be eliminated from this expression to yield rate = k[NO2]^2.

However, I was able to get rate=k[NO2]^2, shown below. Can someone help to explain why what I did was wrong?
since [NO3] is an intermediate, net formation of NO3 = 0 = k1[NO2]^2 - k2[NO3][CO]
---> [NO3] = k1[NO2]^2 / k2[CO]
rate law based on the slow reaction: rate = k2[NO3][CO] = k2 (k1[NO2]^2 / k2[CO]) [CO] = (k1 k2 [NO2]^2 [CO])/(k2 [CO]) = k1[NO2]^2

Thank you!

Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am

### Re: 15.53 (c)

The second step is the slow step, so rate is determined by this step: rate = k[NO3][CO].
And NO3 is formed from step 1, so the rate = k[NO2]^2[CO].
CO cannot be eliminated from the expression since it appears in the slow step.
I do not get why you [NO3] = k1[NO2]^2 / k2[CO].