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### Topic: Problem of the Week - 5/20/10  (Read 9431 times)

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#### azmanam

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##### Problem of the Week - 5/20/10
« on: May 20, 2010, 11:08:56 AM »
Er... problem of the ... indefinite time span?

Sorry for being away so long.  I've been writing up my dissertation and otherwise trying to get the hell out of here (note: it's not working.   I feel like the 'mister! mister!' lady from Happy Gilmore).  Anyway, my labmate and I spent about 15-20 minutes working on this hypothetical, and we were pretty proud of ourselves when we got to the answer.  Thought it would be a fun problem for all to try.

QUESTION:  Suppose a reaction gives you 30% yield and 100% based on recovery of starting material (i.e. 100% mass balance, 30% is product, the only other product off the column is the rest of your recovered starting material).  You want to recycle the recovered starting material by running the same reaction again, recover that unreacted SM, recycle again, recover SM, recycle, ... etc.  derive a formula for predicting the overall cumulative yield over x number of recycles given y % yield for the reaction (and assuming same yield and 100% mass balance each time).  If i didn't explain it well enough, ask and I'll try again
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#### azmanam

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##### Re: Problem of the Week - 5/20/10
« Reply #1 on: May 20, 2010, 11:21:56 AM »
oh, and it'd be nice to show your work and logic leading to your conclusion (including missteps if you're brave enough)
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#### Mitch

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##### Re: Problem of the Week - 5/20/10
« Reply #2 on: May 20, 2010, 11:59:57 AM »
Just to be cheeky, this equation will give you the total unreacted yield in x number of steps.

${ Y_{un} = \text{ e}^{-0.357 x} \text{ X 100%}}$

I wish I could say how I did it. All these years in science has hard-wired my brain to naturally complete first-order differential equations without writing steps.
« Last Edit: May 20, 2010, 01:07:20 PM by Mitch »
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#### sjb

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##### Re: Problem of the Week - 5/20/10
« Reply #3 on: May 20, 2010, 12:17:45 PM »
After 1 reaction, your starting material has "decayed" to 70% original concentration, after n reactions 0.7n remains, so you have created (1 - 0.7n) x 100% of product. Or more generally  (1 - [1 -y]n)  x 100%, for a yield of y% per reaction and n reactions.

Whch is probably the same as Mitch's answer

Similar to extraction formulae, at heart, after all.

#### Schrödinger

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##### Re: Problem of the Week - 5/20/10
« Reply #4 on: May 20, 2010, 12:28:26 PM »
The one reason I don't like Problem of the Week is that even before I start thinking about the problem, the answer is revealed

Anyway, I was going for the same approach as sjb and Mitch. Geometric series summation.
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#### Borek

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##### Re: Problem of the Week - 5/20/10
« Reply #5 on: May 20, 2010, 03:23:21 PM »
I wish I could say how I did it.

They don't give these PhDs for nothing.
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#### Mitch

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##### Re: Problem of the Week - 5/20/10
« Reply #6 on: May 20, 2010, 08:14:43 PM »
I wish I could say how I did it.

They don't give these PhDs for nothing.

Arrrgggg..

${ \frac{dY_\text{un}}{dt} = -\lambda Y_\text{un}}$

${ Y_{un} = Y_{un, initial}e^{-\lambda x}}$

Yun, initial = 100%. At x = 1step Yun/Yun, initial = 0.7

${ ln|0.7| = -\lambda(1) }$

${ \lambda = -0.3567 }$

So the final expression is simply,

${ Y_{un} = \text{ e}^{-0.357 x} \text{ X 100%}}$
« Last Edit: May 20, 2010, 08:33:03 PM by Mitch »
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#### azmanam

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##### Re: Problem of the Week - 5/20/10
« Reply #7 on: May 21, 2010, 07:18:38 AM »
good, good.  Our reasoning was along the lines of sjb, although we approached it a bit differently and came up with a different equation that gets to the same place.  Although, Mitch, I don't understand a word of what you just said

Follow up question.  What if the mass balance isn't 100%?  What if you get 30% product, but only 60% recovered SM?  Do your equations still hold?  Ours does.  Derive an equation for that case.
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#### Mitch

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##### Re: Problem of the Week - 5/20/10
« Reply #8 on: May 22, 2010, 01:32:11 PM »
@azmanam Just open up the first couple pages in your p-chem book's kinetics chapter, my logic flows naturally from first-order kinetics. I am curious what your equation looks like?

The follow up question sounds like a simple second-order differential equation, but I really couldn't just whip it out like I did above, it has been too long since I needed to think that way.
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#### MrTeo

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##### Re: Problem of the Week - 5/20/10
« Reply #9 on: May 23, 2010, 06:48:50 AM »
I think the formula could be made more general starting from sjb's approach and changing the percentage recovered from the reaction this way:

${ sm_*=\frac {sm}{100} \cdot k }$

Where k is the sum of the percentages of products and starting materials recovered (we could call it "mass-balance percentage").
This way we know that after each step we'll have (k⋅sm*)% of reagents avaiable and k(1-sm*)% of product. The only thing we need to consider is that after n steps the total amount of reagents and products is kn (e.g. if k=70% we'll have (0.7)n).

So the final equation should be:

${ p(n)=(1-(1-sm_*)^n) \cdot (k)^n }$
« Last Edit: May 23, 2010, 07:06:56 AM by MrTeo »
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#### azmanam

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##### Re: Problem of the Week - 5/20/10
« Reply #10 on: June 07, 2010, 09:23:07 AM »
sorry, it's been a while. Mitch asked for our equation, so here's our thought process on how we got to it from 30% yield and 70% recovered SM

iteration 1 yield: 1 x 0.3
iteration 2 yield: (1 x 0.7) x 0.3
iteration 3 yield: ((1 x 0.7)0.7) x 0.3
iteration 4 yield: (((1 x 0.7)0.7)0.7) x 0.3
...

thus, total yield for all iterations:

total yield = 0.30[ (1) + (1 x 0.7) + (1 x 0.72) + (1 x 0.73) ]
total yield = 0.30[ (1 x 0.70) + (1 x 0.71) + (1 x 0.72) + (1 x 0.73) ]
total yield = 0.30(0.70 + 0.71 + 0.72 + 0.73)

or

Σx=0→n y(1-y)x
(where y is the yield for each iteration and n is the number of iterations)

or more generally for any recycling experiment where the mass balance is not 100%

Σx=0→n y(z)x
(where y is the yield for each iteration and z is the percent recovered starting material and n is the number of iterations)

There might be a way to convert the summation into a function, but I don't remember my calculus that well.
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