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anonymous
 4 years ago
Nitrogen dioxide decomposes at 300°C via a secondorder 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 2.50L reaction vessel at 300°C. If the halflife and the rate constant at 300°C are 11 seconds and 0.54 M1 s1, respectively, how many moles of NO2 were in the original sample?
I just need an idea on how to approach this problem.
anonymous
 4 years ago
Nitrogen dioxide decomposes at 300°C via a secondorder 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 2.50L reaction vessel at 300°C. If the halflife and the rate constant at 300°C are 11 seconds and 0.54 M1 s1, respectively, how many moles of NO2 were in the original sample? I just need an idea on how to approach this problem.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How about the second order rate law equation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[Rate=k[NO_2]^2\]I have the rate constant, but how do I determine the rate or the concentration of NO_2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, I was talking about this equation\[1 / [A]_{t}^{} = 1/[A]_{0}^{} + kt\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, the integrated rate law for a second order reaction. Ok.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I haven't done this in a year, so I forgot all the terminologies and stuff, sorry.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That gives me an idea of what to do. Let me work a few things out.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[t _{1/2}^{} = 1 / (k [A]_{0}^{})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{[A]_t}=\frac{1}{[A]_0}+kt\]\[[A]_t=\frac{1}{2}[A]_0\]\[\frac{2}{[A]_0}=\frac{1}{[A]_0}+kt\]I then solved for [A]_0 to get 5.94 M^1, and multiplied by the volume of the solution. Thank you!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Seems like I haven't forgotten all my chem yet, woohoo :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Chemistry knowledge is so much more volatile than any other subject :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Volatile or valuable? Chem is my favorite science :) Bio is too much memorization. Physics is too much math, chem is the right balance :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Volatile. 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Chem is still the best, I agree <3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The answer to the question is 0.42mol, if someone happens to be trying to learn from this.
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