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<div class="moz-cite-prefix">On 03/24/2015 08:11 PM, wrote:<br>
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cite="mid:C40B2F181831EF44A88CD73525827803130F1777D7@MERCERMAIL.MercerU.local"
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<p>Dr Pounds,<br>
</p>
<p>I am getting a little confused with how to determine bounds
for the solver. Could you give me some tips?<br>
</p>
<p><br>
</p>
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<font face="serif"><br>
So the first thing is that you need to get Q and K calculated and
them use them to determine which way the reaction has to shift to
establish equilibrium.<br>
<br>
Based on the way the reaction has to shift we have always written
our ICE tables so that our <img style="vertical-align: middle"
src="cid:part1.00090409.04070803@mercer.edu" alt="$x$"> values
will be between 0 and some positive value. Now, lets look at some
examples.<br>
<br>
<img style="vertical-align: middle"
src="cid:part2.02020409.06000501@mercer.edu" alt="$12.4 =
\frac{(2+x)(3+2x)^2}{(4-2x)^2}$"><br>
<br>
First, always look for the parts of the expression where you are
subtracting. Then recognize that the expression in parenthesis
can never be negative. For the example above <img
style="vertical-align: middle"
src="cid:part1.00090409.04070803@mercer.edu" alt="$x$"> can take
on values from 0 up to the open domain 2. Those then would be your
bounds. I then would put an initial value of <img
style="vertical-align: middle"
src="cid:part1.00090409.04070803@mercer.edu" alt="$x$"> into my
solver that is between those two number. In this case I would
enter 1 for <img style="vertical-align: middle"
src="cid:part1.00090409.04070803@mercer.edu" alt="$x$"> and then
solve. Using solver I get <img style="vertical-align: middle"
src="cid:part1.00090409.04070803@mercer.edu" alt="$x$">=0.864.
<br>
<br>
Now, let's look at a more challenging example...<br>
</font><br>
<font face="serif"><img style="vertical-align: middle"
src="cid:part7.09090203.00010404@mercer.edu" alt="$12.4 =
\frac{(3+2x)^2}{(4-2x)^2(5-3x)^3}$"><br>
<br>
This is more difficult because I now have to look at TWO
expressions to decide how to set my bounds. Remember, the stuff
in parenthesis can't be zero.<br>
<br>
For <img style="vertical-align: middle"
src="cid:part8.04040500.03040304@mercer.edu" alt="$(4-2x)$"> the
biggest value <img style="vertical-align: middle"
src="cid:part1.00090409.04070803@mercer.edu" alt="$x$"> can be
is 2.<br>
<br>
For <img style="vertical-align: middle"
src="cid:part10.01060206.07040702@mercer.edu" alt="$(5-3x)$">
the biggest value of <img style="vertical-align: middle"
src="cid:part1.00090409.04070803@mercer.edu" alt="$x$"> can be
is 5/3, or 1.667. We MUST use the smaller of these (1.667)
because anything bigger would make the other expression negative.
So in this case you would set your bounds to 0 and 1.667 and
select an initial value of <img style="vertical-align: middle"
src="cid:part1.00090409.04070803@mercer.edu" alt="$x$"> between
those two numbers. I chose 1 again for mine. Solver in this
case give <img style="vertical-align: middle"
src="cid:part1.00090409.04070803@mercer.edu" alt="$x$">=1.3<br>
<br>
Does that help?<br>
<br>
<br>
</font>
<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>
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