anonymous 4 years ago Nitrogen dioxide decomposes at 300°C via a second-order process to produce nitrogen monoxide and oxygen according to the following chemical equation. 2 NO2(g) → 2 NO(g) + O2(g). A sample of NO2(g) is initially placed in a 2.50-L reaction vessel at 300°C. If the half-life and the rate constant at 300°C are 11 seconds and 0.54 M-1 s-1, respectively, how many moles of NO2 were in the original sample? I just need an idea on how to approach this problem.

1. anonymous

How about the second order rate law equation?

2. anonymous

$Rate=k[NO_2]^2$I have the rate constant, but how do I determine the rate or the concentration of NO_2?

3. anonymous

Well, I was talking about this equation$1 / [A]_{t}^{} = 1/[A]_{0}^{} + kt$

4. anonymous

Oh, the integrated rate law for a second order reaction. Ok.

5. anonymous

I haven't done this in a year, so I forgot all the terminologies and stuff, sorry.

6. anonymous

That gives me an idea of what to do. Let me work a few things out.

7. anonymous

$t _{1/2}^{} = 1 / (k [A]_{0}^{})$

8. anonymous

$\frac{1}{[A]_t}=\frac{1}{[A]_0}+kt$$[A]_t=\frac{1}{2}[A]_0$$\frac{2}{[A]_0}=\frac{1}{[A]_0}+kt$I then solved for [A]_0 to get 5.94 M^-1, and multiplied by the volume of the solution. Thank you!

9. anonymous

Seems like I haven't forgotten all my chem yet, woohoo :)

10. anonymous

Chemistry knowledge is so much more volatile than any other subject :P

11. anonymous

Volatile or valuable? Chem is my favorite science :) Bio is too much memorization. Physics is too much math, chem is the right balance :D

12. anonymous

Volatile. Chemistry is so much based on your understanding of the subject rather than the pure math behind it. If you don't fully understand the subject (like I don't fully understand rate laws), it will disappear in a matter of weeks. I can not do physics for years, and just jump right back in, because it's just basic problem solving and algebra :P

13. anonymous

Chem is still the best, I agree <3

14. anonymous

The answer to the question is 0.42mol, if someone happens to be trying to learn from this.