<html>
<head>
<meta http-equiv="content-type" content="text/html; charset=UTF-8">
</head>
<body>
<p><font face="serif">I have had severa people ask how to find the
activation energy in the temperature kinetics lab. <br>
</font></p>
<p><font face="serif">In the lab procedures you are give the
equation</font></p>
<p><font face="serif"><math
xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mo
lspace="0em" rspace="0em">ln</mo><mo stretchy="false">(</mo><mi>R</mi><mi>A</mi><mi>T</mi><mi>E</mi><mo
stretchy="false">)</mo><mo>=</mo><mo>−</mo><mfrac><msub><mi>E</mi><mi>a</mi></msub><mi>R</mi></mfrac><mfrac><mn>1</mn><mi>T</mi></mfrac><mo>+</mo><mo
lspace="0em" rspace="0em">ln</mo><mo stretchy="false">(</mo><mi>c</mi><mi>o</mi><mi>n</mi><mi>s</mi><mi>t</mi><mi>a</mi><mi>n</mi><mi>t</mi><mo
stretchy="false">)</mo></mrow><annotation encoding="TeX">\ln(RATE)=-\frac{E_a}{R}\frac{1}{T}+\ln(constant)</annotation></semantics></math><br>
</font></p>
<p><font face="serif">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
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>−</mo><mfrac><msub><mi>E</mi><mi>a</mi></msub><mi>R</mi></mfrac></mrow><annotation
encoding="TeX">-\frac{E_a}{R}</annotation></semantics></math><br>
</font></p>
<p>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.</p>
<p>Let me know if you have any questions.</p>
<p><br>
</p>
<div class="moz-signature">-- <br>
<b><i>Andrew J. Pounds, Ph.D.</i></b><br>
<i>Professor of Chemistry and Computer Science</i><br>
<i>Director of the Computational Science Program</i><br>
<i>Mercer University, Macon, GA 31207 (478) 301-5627</i></div>
</body>
</html>