[CHM 112] Finding the Activation Energy

[Andrew J. Pounds]

I have had severa people ask how to find the activation energy in the 
temperature kinetics lab.

In the lab procedures you are give the equation

ln(RATE)=−EaR1T+ln(constant)\ln(RATE)=-\frac{E_a}{R}\frac{1}{T}+\ln(constant)

If you plot the log of the rates on the y-axis and the values of 1/T 
(where T is in Kelvin) on the x-axis then the slope of the line should 
be equal to −EaR-\frac{E_a}{R}

The slope will have units of T.   If you multiply the slope by -R (which 
should be 8.314 J/deg-mol) then you should get a positive activation 
energy with units of J/mol.

Let me know if you have any questions.


-- 
*/Andrew J. Pounds, Ph.D./*
/Professor of Chemistry and Computer Science/
/Director of the Computational Science Program/
/Mercer University, Macon, GA 31207 (478) 301-5627/
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