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<p><font face="serif">Hey -- let's all make sure that we do this
right... review the lab description and then read the
following...<br>
</font></p>
<p><font face="serif">The Arrhenius equation is</font></p>
<p><font face="serif"><br>
</font></p>
<p><font face="serif"><math display="block"
xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>k</mi><mo>=</mo><mi>A</mi><msup><mi>e</mi><mrow><mo>−</mo><mfrac><msub><mi>E</mi><mi>a</mi></msub><mrow><mi>R</mi><mi>T</mi></mrow></mfrac></mrow></msup></mrow><annotation
encoding="TeX">k = A e^{-\frac{E_a}{RT}}</annotation></semantics></math><br>
</font></p>
<p><font face="serif">In this lab the concentrations are all the
same, so the rate is directly proportional to the rate; we are
essentially replacing the rate constant in the Arrhenius
equation with the rate.<br>
</font></p>
<p><font face="serif">We take the natural log of both sides to get
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>l</mi><mi>n</mi><mo
stretchy="false">(</mo><mi>rate</mi><mo stretchy="false">)</mo><mo>=</mo><mi>l</mi><mi>n</mi><mo
stretchy="false">(</mo><mi>A</mi><mo stretchy="false">)</mo><mo>−</mo><mfrac><msub><mi>E</mi><mi>a</mi></msub><mi>R</mi></mfrac><mfrac><mn>1
</mn><mi>T</mi></mfrac></mrow><annotation encoding="TeX">ln(k)
= ln(A) - \frac{E_a}{R} \frac{1}{T}</annotation></semantics></math><br>
</font></p>
<p><font face="serif">Notice how this looks like the equation of a
line: y = b + mx</font></p>
<p><font face="serif">So you plot the natural log of rate on the y
axis and 1/T on the x-axis. Make sure T is in kelvin. Since
your rates will all be less than 1 (but not negative) the
natural log of your rates will be negative. You should get a
graph with a negative slope. That is okay, however, because
that negative slope is equal to
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>−</mo><msub><mi>E</mi><mi>a</mi></msub><mo>/</mo><mi>R</mi></mrow><annotation
encoding="TeX">-E_a/R</annotation></semantics></math>.
The negatives cancel and you should be left with a positive
activation energies.</font></p>
<p><font face="serif">If I see negative activation energies reported
bad things will happen...<br>
</font></p>
<p><font face="serif"> </font><br>
</p>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Director of the Computational Science Program
Mercer University, Macon, GA 31207 (478) 301-5627
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