NO2 decomposes into NO and O2.

**1. ****At high temperatures, nitrogen dioxide, ***NO*2, decomposes into *NO *and *O*2. If *y*(*t*) is the concentration of *NO*2 (in moles per liter), then at 600º*K*, *y*(*t*) changes according to the reaction law *dy*/*dt *= −.05*y*2 for time *t *in seconds.

**A. ****Express ***y *in terms of *t *and the initial concentration *y*o.

**Equation of y in terms of t and initial concentration y****o**

*Solve for constant of integration.*

Solve for differential equation.

Hello, I can't seem to find the proper method to set up this problem. The chem. stuff is slightly misleading and I can't seem to come up with anything satisfying. **Any help on setting this up would be appreciated! **

The one below is merely an extension of the aforementioned question, when I figure out the one above it shouldn't be too difficult.

**B. ****Assuming that the concentration of ***NO*2 is twice as high at *t *= 20 seconds as it is at 100 seconds, find the exact initial concentration of the *NO*2. Reminder: “Exact” means no calculator numbers.

**Exact initial concentration y****o**

Re: Need help on setting up differential equation.

$\displaystyle \frac{dy}{dt}=-\frac{y^2}{20}\Rightarrow \frac{dy}{y^2}=-\frac{dt}{20}$

Re: NO2 decomposes into NO and O2.

Thanks!

I had a similar setup but I made some err on the orientation. Moreover I was wondering as to how this equation can be set up to include Y sub-zero(the initial concentration) and again how to utilize the temperature so nicely conceded in the problem. I'm having issue's weaving all this together (Headbang).

Re: NO2 decomposes into NO and O2.

You use $\displaystyle y_0$ to find your constant of integration since you have the ordered pair $\displaystyle (0, y_0)$

Re: NO2 decomposes into NO and O2.

When I try solving for dy/dt=-.05y^2 I yield 20/Y=t. And isolating for Y I get Y(x)=20/t + the constant. However in this situation I can't plug in t=0 as it would be undefined. I'm probably messing up somewhere, thanks for your guys patience!

Re: Need help on setting up differential equation.

{ Quote:

Originally Posted by

**dwsmith** $\displaystyle \frac{dy}{dt}=-\frac{y^2}{20}\Rightarrow \frac{dy}{y^2}=-\frac{dt}{20}$

$\displaystyle \int\frac{dy}{y^2}=-\frac{1}{20}\int dt\Rightarrow \frac{-1}{y^1}=\frac{-t}{20}+C\Rightarrow y^{-1}=\frac{t}{20}+C_2$

$\displaystyle \Rightarrow y(t)=\frac{1}{\frac{t}{20}+C_2}\Rightarrow y(t)=\frac{20}{t+C_3}$

Re: Need help on setting up differential equation.

Thank You! (Bow) I was able to figure it out thanks to the help of both of you. It was rather intuitive after solution, but that's probably hindsight bias. Thanks again!

Re: Need help on setting up differential equation.

I'm sorry for necroing an old thread but I am currently working on the same problem and was also thrown off by the chemistry presented in the problem and I was able to follow but do they want the y_{0} to come from y(0) = 1/(.05t + C) and solve for C yielding C = 1/y_{0} so you would end up with y(t) = 1/(.05t + 1/y_{0})