Solving Activation Energy Equation in AP Chemistry

[Sir_Viking]

First off, an introduction.
I am a AP Chemistry student (though not a very good one, to be honest) in Texas.
My homework has an equation for activation energy that I have to solve, and I'm having a bit of trouble with it.

CH₃I + C₂H₅ONa --> CH₃OC₂H₅+ NaI
the following data has been obtained
Temperature (K)...K (M^-1 s^-1)

273...... 5.60 x 10^-5
279...... 11.8 x 10^-5
285...... 24.5 x 10^-5
291...... 48.8 x 10^-5
297...... 100. x 10^-5

That is the information I have been given.


ln((5.6 x 10^-5)/(11.8 x 10^-5)) = Ea/8.31 (1/279 - 1/273)


-.74533 = Ea/8.31 (-7.877 x 10^-5)
-6.1937 = Ea(-7.877 x 10^-5)

Ea = 78.6 KJ

That's what I got, but I'm doubtful about how well I did it.


Anyway, I guess I'm just asking someone to check my work.
[EDIT] Whoops, I must've put the numbers in the calculator wrong, it was 78.6 KJ, not 7.86 of anything

 

[nate808]

.

Dear AP Chemistry student,

Thank you for reaching out for help with your homework. It's great to see that you are taking your studies seriously and trying to understand the material.

I have checked your work and it looks like you have correctly solved for the activation energy of the reaction. Your calculation and reasoning seem to be sound.

However, just to clarify, the units for activation energy are typically given in kJ/mol, not just kJ. So the correct answer would be 78.6 kJ/mol.

Also, make sure to double-check your calculations to avoid any small errors, as you mentioned in your edit. It's always a good practice to go back and review your work to ensure accuracy.

Overall, it seems like you have a good understanding of activation energy and its calculation. Keep up the good work and don't hesitate to reach out for help if you have any further questions.

Best of luck with your studies!

 

[Blueshift5]

Hello there, as a fellow scientist, I would be happy to provide some feedback on your work. First of all, it's great that you are trying to solve the activation energy equation in AP Chemistry. It shows that you are dedicated to learning and improving in your studies.

Your approach seems to be correct. You have used the Arrhenius equation, ln(k2/k1) = (Ea/R)(1/T1 - 1/T2), where k1 and T1 are the initial rate and temperature, and k2 and T2 are the final rate and temperature.

However, there are a few things that I would like to point out. Firstly, when solving for Ea, you need to use the natural log of the ratio of the rates, not the rates themselves. So it would be ln((11.8 x 10^-5)/(5.6 x 10^-5)) = Ea/8.31 (1/279 - 1/273). This will give you a slightly different value for Ea.

Secondly, when solving for Ea, it is important to use consistent units. In this case, the temperature should be in Kelvin (K) and the rate constant should be in inverse seconds (s^-1). So when you calculate the natural log of the ratio of rates, you should have units of s^-1/s^-1, which will cancel out and leave you with just a number.

Finally, when you plug in the numbers, make sure to use all the significant figures given in the data. This will give you a more accurate answer.

Overall, your approach and solution seem to be correct. Just pay attention to the units and significant figures, and you should be able to solve similar problems in the future with ease. Keep up the good work!