[Nav]

Hello all,

I am a first year university student with a real noodle-scratcher of a question. We have just started Chemical Kinetics and they threw this question at us first. Here goes:

The rate constant, k, for a given reaction was determined at a number of temperatures. A graph of log k versus the reciprocal of the Kelvin temperature yielded a straight line with a slope of -250 K. What is the activation energy of this reaction in KJ/mol?

I have searched through the book and asked several classmates as to how this problem should be tackled but to no avail. I thought that if I had the equation, I could find the y-intercept but I'm not sure what that would reveal. Any help would be greatly appreciated. Thanks in advance

Regards,

Nav

[Borek]

Arrhenius law

[Nav]

Hmm, at first it seems to be the key. But the reaction is unknown, so I cannot determine an appropriate A value. However, I did come across this link:

http://members.aol.com/logan20/tempefct.html

It states that the slope is simply the activation energy over R, the universal gas law. If you multiply the slope ( -250K) by 8.31 X 10^-3 KJ/mol then the Kelvin's cancel out, leaving only KJ/mol, the proper units of activation energy. Does this make sense or have i overlooked something?

Thanks again.

[Yggdrasil]

Correct.  Here's the explanation based on the Arrhenius equation.

The Arrhenius equation states that the rate constant, k, of a chemical reaction depends exponentially on the activation energy, E_a:

k = Ae^-E_a/RT

where A, the pre-factor, depends on the specifics of the reaction that occurs.  Now, take the natural log of both sides of the equation:

ln k = ln (Ae^-E_a/RT) = ln A - E_a/RT

Now let ln k = y and 1/T = x.  Now the equation becomes:

y = ln A - (E_a/R)x

Thus, when you plot ln k versus 1/T, you get a line with slope -E_a/R and a y-intercept of ln A.  So, from your graph, not only can you figure out E_a, but you can also get an idea of the pre-factor as well.

[champ]

If you look at Page #594(General Chemistry (Ninth Edition)) and at the really end you will see "To Evaluate Ea" the slope is given and you have to find the Ea.
In this question the slope is given which is -250K. The Equation is gonna be Slope=-Ea/R if u cancel out the units you will end up with J/mol convert J/mol to kJ/mol divided my 1000 and the answer i got is 2.08kJ/mol
If i am wrong then please Correct me Thank you very much!!!

[champ]

 read more about Activation Energy...
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/activation.htm
Thank very much!!

[Yggdrasil]

2.08kJ/mol looks like a good answer to me.

[Nav]

Thank you very much for all the replies! I really appreciate it

[cinatas]

A question if I may (or an observation):

Didn't the question mention that it was plotting log k as opposed to ln k? doesn't that mean that 2.08kJ/mol should be multiplied by 2.303 (ie. ln 10) in order to convert the natural logarithm to a logarithm of base 10?