## NO2 Example from Lecture

Ava Harvey 2B
Posts: 54
Joined: Fri Sep 29, 2017 7:04 am

### NO2 Example from Lecture

I was a bit confused in lecture today during the example that Lavelle was walking us through regarding the chemical reaction 2NO2 --> 2NO + O2. I understand the concept that the concentration of NO2 is decreasing at a rate that is equal to the rate at which the concentration of NO is increasing (as the stoichiometric coefficients are the same). I also understand that the concentration of O2 is increasing at a rate that is half as fast compared to the rate of NO2. However, I was confused as to why we multiply the rate of concentration for [O2] by 2 when writing the equation -d[NO2]/dt = d[NO]/dt = 2 * d[O2]/dt. From my understanding, the concentration of O2 is increasing half as fast, so why wouldn't it be multiplied by 1/2 and not 2? If someone wouldn't mind explaining this to me that would be great! Thanks so much!

Lisa Tang 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:05 am
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### Re: NO2 Example from Lecture

The equation is -d[NO2]/dt = d[NO]/dt = 2 * d[O2]/dt because as you stated, the rate at which NO2 concentration decreases is equal to the rate at which NO concentration increases. In addition, I believe that because O2 concentration increases half as fast, 2 x the rate of oxygen concentration increase equals the other rates.
It makes sense because by multiplying the rate by 2, it states that the rate of formation of O2 is half the rate of consumption of NO2 and NO; the rate of formation of O2 is slower.
If the rate was multiplied by 1/2, it would instead mean that the rate of formation of O2 is twice the rate of consumption of the reactants (which is not the case in this example).
I hope this helps.

Julie Steklof 1A
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

### Re: NO2 Example from Lecture

The O2 concentration increases half as fast as the NO2 concentration decreases. For these rates to be equal, d[O2]/dt must be multiplied by two so it equals -d[NO2]/dt.