<html>
<head>
<meta content="text/html; charset=ISO-8859-1"
http-equiv="Content-Type">
</head>
<body bgcolor="#FFFFFF" text="#000000">
<div class="moz-cite-prefix">On 07/04/13 22:17, wrote:<br>
</div>
<blockquote
cite="mid:C40B2F181831EF44A88CD735258278030261C41AFF@MERCERMAIL.MercerU.local"
type="cite">
<meta http-equiv="Content-Type" content="text/html;
charset=ISO-8859-1">
<style id="owaParaStyle"><!--P {
        MARGIN-BOTTOM: 0px; MARGIN-TOP: 0px
}
--></style>
<div style="direction: ltr;font-family: Tahoma;color:
#000000;font-size: 10pt;">
<p>Hey Dr. Pounds,</p>
<p>On Chapter 14 #44, it is asking us to find Temperature 2. I
thought I knew how to do it, but I think my algebra is messing
up somewhere. On your solutions, you got 644 K, but I keep
getting a tiny number. This is my work...</p>
<p> </p>
<p>ln(k2/k1)=(Ea/R)[(1/T1)-(1/T2)]</p>
<p> </p>
<p>ln(8.80E-4/4.60E-4)= (104000J/8.314 J)[(1/623)-(1/T2)]</p>
<p> </p>
<p>.648695418= 12509.02093[(1/623)-(1/T2)]</p>
<p> </p>
<p>5.18582E-5= [(1/623)-(1/T2)]</p>
<p> </p>
<p>.0015532782= (1/T2)</p>
<p> </p>
<p>THIS IS WHERE I AM GETTING LOST. I am not too sure if I did
my math wrong or did the whole problem wrong. Sorry this was
so lengthy!
</p>
<p> </p>
<p> </p>
<p>Thank you!</p>
</div>
</blockquote>
<br>
<br>
<font face="serif">You are so close that I a am wondering why you
stopped -- you need to find <img style="vertical-align: middle"
src="cid:part1.02070403.03090103@mercer.edu" alt="$T_2$">...<br>
<br>
<img style="vertical-align: middle"
src="cid:part2.04070600.05010701@mercer.edu"
title="\documentclass{article}
\usepackage[utf8x]{inputenc}
\pagestyle{empty}
\begin{document}
$\mathrm{0.0015532782} = \frac{1}{T_2}$ %this is where your
LaTeX expression goes
\end{document}
" alt="\documentclass{article}
\usepackage[utf8x]{inputenc}
\pagestyle{empty}
\begin{document}
$\mathrm{0.0015532782} = \frac{1}{T_2}$ %this is where your
LaTeX expression goes
\end{document}
"></font><br>
<br>
<img style="vertical-align: middle"
src="cid:part3.05080605.06090003@mercer.edu" alt="$T_2 =
\frac{1}{\mathrm{0.0015532782}}$"><br>
<br>
<img style="vertical-align: middle"
src="cid:part4.04050708.06040100@mercer.edu"
alt="$T_2=\mathrm{644}$"><br>
<br>
That's your temperature in Kelvin -- and the correct answer.<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>
</body>
</html>