Rate law question.. concerning nitrogen dioxide?

Question

nitrogen dioxide decomposes to nitrogen monoxide and oxygen:

2NO_2 --> 2NO + O_2 (all in gaseous state)

at what rate will O-2 concentration change? I think the answer is half as fast as NO gas. Correct?

If you increase the temperature of the reaction from 300 degrees Celsius to 400 degrees Celsius, how will this affect the rate of decomposition of NO_2? -- For this one I'm not sure. It seems that it would decompose more quickly, since the molecules will collide more frequently with their increased average kinetic energy. Is this correct? Is there a specific formula to illustrate this?

This is another part of the problem: "Your friend looks at the balanced equation and writes the rate law for the reaction: Rate = k[NO_2]^2 You caution your friend that you can't just look at the overall balanced equation to determine the rate law. How could you determine whether your friend is correct?

? Thanks for the help. 8^?

Answer 1

I'm pretty sure you're right with the first one... the O2 should change half as fast as the NO2 and the NO.

For the second, if you increase the temperature, the rate of decomposition will increase, for exactly the reason you said. Umm... there's a really, really ugly formula to go along with it that looks like...

ln(rate 1/rate 2) = -(Activation Energy/ the gas constant 8.31 J/mol)((1/ temp 1)-(1/temp 2) + ln (a given value).

For the third part, you can only determine that your friend is correct through the use of experimental data. For example, if you're given a table that says for a certain amount of NO2, you get a certain rate of appearance of NO, you can determine the rate law, but otherwise.... no good!
However (and this is probably more info than you need), for intermediate steps, the ones that have products that'll cancel out by the end, you can use your coefficients to determine the rate law.