How to Calculate Argon Atom Energy Fractions and Reaction Rates?

[erok81]

I am completely stuck on a couple of questions and would like a bit of help on these.

First one. This I have no idea where to even start and I can't find anything on it in my text either. So maybe a formula or point me in the right direction would help.

Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 12.5 kJ or greater.

This one is a bit easier. I am not 100% sure on how to do it, but I did guess this one right.

The question is:

The activation energy of a certain reaction is 76.7 kJ/mol. How many times faster will the reaction occur at 50°C than at 0°C, assuming equal initial concentrations of reactants?

I used this equation:

ln k = ln A -\frac {E_a}{RT}

I did that for both temps then divided them. I got a close answer to the correct one. The thing I am confused on is the A. My book says that is for the frequency factor when using the Arrhenius equation. I read in Wikipedia that is just the sqrt of the temp. So I did that and it came out close, but I am not sure if that is right. So I really just need to know how you find A.


Thanks for the help.

 

[ksinclair13]

We didn't cover this when I learned about kinetics a few weeks ago in AP Chemistry. Therefore, I would wait for an expert to help you out on this, but you may want to take my thoughts into consideration in case I am right.

Use this derived equation:

ln(\frac{k_2}{k_1}) = - \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})

My answer came out to the reaction rates being approximately equal. This is exactly what I did:

ln(x) = - \frac{76.7}{8.3145} (\frac{1}{273} - \frac{1}{323})

ln(x) = -9.22 (.000567)

ln(x) = -.00523

e^{ln(x)} = e^{-.00523}

x = .995

\frac{k_2}{k_1} = .995

This means that the rate at 50°C is about .995 the rate at 0°C.

Once again, do not take my word on this for I did learn any of this in my class.

 

[erok81]

That looks like it makes sense, but the answer I guessed was 187, and it was right. But I am pretty sure I got it wrong. I ended up with an answer that was close using the equation I posted, so I chose it and was right.

But, while we are on the subject of using the equation you posted, if you have a second, enter this problem into it.

"The rate constant for a reaction is 1.5 x 10-2 s-1 at 775 K and 3.5 x 10-2 s-1 at 825 K. What is the activation energy?"

That is the one I know how to do, but no matter how I enter into that equation you posted, I can't get the right answer.

 

[t!m]

Use this derived equation:

ln(\frac{k_2}{k_1}) = - \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})

You have this equation a little fudged. It should read:

ln(\frac{k_2}{k_1}) = - \frac{E_a}{R} (\frac{1}{T_2} - \frac{1}{T_1})

Additionally, you have to be mindful of converting kilojoules into joules, sinclair.

This is the same expression you posted erok, just subtracted from itself.
Using the above equation and watching your units, you'll arrive at the expected 187.

 

[erok81]

Ah ok. That makes sense, I have no clue how I ended up with the right answer, since I used something totally different.

Same with post #3, I did it again this morning waiting for class to start and came up with the right answer.


Any clues on what formulas to use on that first one?

 

[siddharth]

For the first one, have you learned about Boltzmann distribution?

 

[erok81]

For the first one, have you learned about Boltzmann distribution?

I don't think we have this semester. I know last semester chem we did.

The main problem is, I was sick and missed a day of class, and I think they went over whatever we are supposed to do then. Because I can't find anything in book in the chapter we are covering right now.

 

[ksinclair13]

I had good feeling that I was wrong. I was just trying my best to point him in the right direction (it sort of worked ). That is why I said not to trust what I said . I actually did figure it out using the correct equation first (1/T2 - 1/T1) and got 1.xx, which makes more sense now. If I would've just converted into joules, I think I would've gotten it correct. Sorry.

 

[siddharth]

These links may help you

http://www.webchem.net/notes/how_far/kinetics/maxwell_boltzmann.htm
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/disfcn.html
http://en.wikipedia.org/wiki/Maxwell-Boltzmann_distribution

 

[erok81]

Hmm..I can't find anywhere on there how you would find the fraction of atoms like in problem one.

 

[siddharth]

The fraction of atoms (\frac{n_i}{N}), occupying a state which has energy E_i is given by

\frac{n_i}{N} = \frac{e^{-\beta E_i}}{\sum_{i} e^{-\beta E_i}}

and where,

\beta = \frac{1}{kT}
where k is the Boltzmann constant.

so, if you know the energy gap between the energy levels, you can find out the fraction.