NO2 decomposes into NO and O2.

**Senaru** · June 13th 2011, 02:41 PM

1. At high temperatures, nitrogen dioxide, NO2, decomposes into NO and O2. If y(t) is the concentration of NO2 (in moles per liter), then at 600ºK, y(t) changes according to the reaction law dy/dt = −.05y2 for time t in seconds.

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

Equation of y in terms of t and initial concentration yo
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 NO2 is twice as high at t = 20 seconds as it is at 100 seconds, find the exact initial concentration of the NO2. Reminder: “Exact” means no calculator numbers.

Exact initial concentration yo

**dwsmith** · June 13th 2011, 04:11 PM

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

**Senaru** · June 13th 2011, 05:06 PM

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

.

**e^(i*pi)** · June 13th 2011, 05:16 PM

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

**Senaru** · June 13th 2011, 05:43 PM

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!

**dwsmith** · June 13th 2011, 06:56 PM

{

Originally Posted by dwsmith

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

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

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

**Senaru** · June 13th 2011, 07:29 PM

Thank You!

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!

**ghoulmaster** · May 1st 2012, 07:33 PM

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 y₀ to come from y(0) = 1/(.05t + C) and solve for C yielding C = 1/y₀ so you would end up with y(t) = 1/(.05t + 1/y₀)

Thread: NO2 decomposes into NO and O2.

LinkBack

Thread Tools

Display

NO2 decomposes into NO and O2.

Re: Need help on setting up differential equation.

Re: NO2 decomposes into NO and O2.

Re: NO2 decomposes into NO and O2.

Re: NO2 decomposes into NO and O2.

Re: Need help on setting up differential equation.

Re: Need help on setting up differential equation.

Re: Need help on setting up differential equation.

Search Tags