[CHM 112] (no subject)

[Andrew J. Pounds]

On 03/24/2015 08:11 PM, wrote:
>
> Dr Pounds,
>
> I am getting a little confused with how to determine bounds for the 
> solver. Could you give me some tips?
>
>

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.

Based on the way the reaction has to shift we have always written our 
ICE tables so that our $x$ values will be between 0 and some positive 
value.  Now, lets look at some examples.

$12.4 = \frac{(2+x)(3+2x)^2}{(4-2x)^2}$

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 $x$ 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 $x$ into my solver that is between those two 
number.  In this case I would enter 1 for $x$ and then solve.     Using 
solver I get $x$=0.864.

Now, let's look at a more challenging example...

$12.4 = \frac{(3+2x)^2}{(4-2x)^2(5-3x)^3}$

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.

For $(4-2x)$ the biggest value $x$ can be is 2.

For $(5-3x)$ the biggest value of $x$ 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 $x$ between those two 
numbers.   I chose 1 again for mine.  Solver in this case give $x$=1.3

Does that help?


-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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