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<div class="moz-cite-prefix">On 07/05/14 16:14, wrote:<br>
</div>
<blockquote
cite="mid:C40B2F181831EF44A88CD73525827803130774D6F5@MERCERMAIL.MercerU.local"
type="cite">
<pre wrap="">Hello Dr.Pounds,
For the additional chat. 14 problems number 5, what is the formula you used for the overall order of reaction?
And this is different than the book example, where it showed us how to divide 2 by 1 and 3 by 2 to get ratios.
Could you please explain further?
Thank you</pre>
</blockquote>
<br>
As I said in class, my methods differ from the book when it comes
to finding the orders of reactions when one is given data. I tend
to use logarithms to solve for the orders with respect to different
components.<br>
<br>
For example, in the additional problem 5 that you mention, the rate
law is going to be something like<br>
<br>
if A = <img style="vertical-align: middle"
src="cid:part1.04080508.08090901@mercer.edu"
alt="$\mathrm{HgCl_2}$"> and B is <img style="vertical-align:
middle" src="cid:part2.07080505.05040508@mercer.edu"
alt="$\mathrm{C_2O_4^{2-}}$"> then the rate law looks like<br>
<br>
<img style="vertical-align: middle"
src="cid:part3.04050404.06090104@mercer.edu" alt="$\mathrm{rate =
k [A]^m [B]^n}$"><br>
<br>
To find the order with respect to <img style="vertical-align:
middle" src="cid:part4.03000500.01080902@mercer.edu"
alt="$\mathrm{[HgCl_2]}$"> I have to use the concentrations and
rates for two experiments where the concentration of <img
style="vertical-align: middle"
src="cid:part4.03000500.01080902@mercer.edu"
alt="$\mathrm{[HgCl_2]}$"> was changing but the concentration of <img
style="vertical-align: middle"
src="cid:part2.07080505.05040508@mercer.edu"
alt="$\mathrm{C_2O_4^{2-}}$"> was constant (experiments 2 and 3 in
this case). Similarly, to find the order with respect to <img
style="vertical-align: middle"
src="cid:part7.01000108.02040107@mercer.edu"
alt="$\mathrm{[C_2O_4^{2-}]}$"> I have to use experiments where
the <img style="vertical-align: middle"
src="cid:part2.07080505.05040508@mercer.edu"
alt="$\mathrm{C_2O_4^{2-}}$"> concentration was changing and the <img
style="vertical-align: middle"
src="cid:part4.03000500.01080902@mercer.edu"
alt="$\mathrm{[HgCl_2]}$"> was constant. I chose to use
experiments 3 and 4 (but I could have also used experiments 1 and
2).<br>
<br>
<br>
<br>
The derivation of this method is found on page 20-21 of the lecture
slides and should also be applicable to the book problems.<br>
By taking a ratio of the rates<br>
<br>
<img style="vertical-align: middle"
src="cid:part10.07030904.01000704@mercer.edu"
alt="$\mathrm{\frac{rate_1=k [A]_1^m [B]_1^n}{rate_2=k [A]_2^m
[B]_2^n}}$"><br>
<br>
and for example, using a case where the <img style="vertical-align:
middle" src="cid:part11.06080601.06030907@mercer.edu"
alt="$\mathrm{[B]_1}$"> and <img style="vertical-align: middle"
src="cid:part12.00080702.04060307@mercer.edu"
alt="$\mathrm{[B]_2}$"> concentrations are the same (and also
noting that the rate constant, k, has to be the same in both
equations) the ratio equation simplifies to:<br>
<br>
<img style="vertical-align: middle"
src="cid:part13.07070906.07090106@mercer.edu"
alt="$\mathrm{\frac{rate_1=[A]_1^m }{rate_2= [A]_2^m}}$"><br>
<br>
Mathematically this is equvalent to <br>
<br>
<img style="vertical-align: middle"
src="cid:part14.09030307.04030903@mercer.edu"
alt="$\mathrm{\left(\frac{rate_1}{rate_2}\right)=\left(\frac{[A]_1}{[A]_2}\right)^m}$"><br>
<br>
Using the fact that <img style="vertical-align: middle"
src="cid:part15.07030006.09030102@mercer.edu" alt="$a=b^n$"> is
the same as <img style="vertical-align: middle"
src="cid:part16.07090906.00050603@mercer.edu" alt="$\ln a = n \ln
b$"> , we can rewrite the equation as<br>
<br>
<img style="vertical-align: middle"
src="cid:part17.02020405.01090400@mercer.edu" alt="$\mathrm{\ln
\left(\frac{rate_1}{rate_2}\right)=m \ln
\left(\frac{[A]_1}{[A]_2}\right)}$"><br>
<br>
or, upon rearrangement<br>
<br>
<img style="vertical-align: middle"
src="cid:part18.06090809.03080107@mercer.edu"
alt="$\frac{\mathrm{\ln \left(\frac{rate_1}{rate_2}\right)}}{\ln
\left(\frac{[A]_1}{[A]_2}\right)}=m }$"><br>
<br>
I repeat the procedure to solve for the order with respect to the
other component and then solve for <img style="vertical-align:
middle" src="cid:part19.03060003.00030906@mercer.edu" alt="$k$">.<br>
<br>
If you find a problem in the book that is set up this same way and
cannot be solved with this method then I would like to know about
it.<br>
<br>
<br>
<br>
<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
Mercer University, Macon, GA 31207 (478) 301-5627
<a class="moz-txt-link-freetext" href="http://faculty.mercer.edu/pounds_aj">http://faculty.mercer.edu/pounds_aj</a>
</pre>
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