Guggenhiem method. Any chemists out there?

[warramungas]

Hi guys.
Well off topic but I was hoping there might be a chemist (or similar) out there that can explain this better than the lecturer I'm trying to work out can or they may know a decent textbook that describes it better. Spent all day Saturday and Sunday trying to work it out and banging my head against the wall. :angry-banghead:
My current unit I'm completing requires me to understand the Guggenheim method for analyzing data to work out the order of reaction. It seems to be heavily reliant on an excel type program to use it.
I usually work out the order of reaction and equation of the line pertaining to particular reactants using other more mathematical methods, depending on given information, but apparently I 'NEED' to know this one and it could be in the exam in some way. Hate it when they try to teach you something you'll probably never use but will test you on it anyway.
However his description on how to do it is vague at best and the six bajillion (6 really) chemistry textbooks I have don't refer to it in any sort of analytical way and my best mate google is keeping it to himself.
Yes I've tried to contact the lecturer and the tutor but the lecturer just repeats what's already been said in the lectures a little louder and the tutor hasn't replied since last Thursday so must be busy. Being an external student my resources are limited as most of the ex uni people I work with don't seem to remember much from uni anyway.
Anybody out there have a deep and meaningful understanding of it?
If not, all good as I'll keep chasing it down. I just hoped someone here might already 'get it'.
Feel free to PM me if you like.
Cheers,
Warra

[db1979]

It doesn't ring a bell but I'll see what I can dig up.

[Doubleuj]

Just googled it, and now my head hurts...

[db1979]

I looked it up, found a site that likes to explain chemistry with maths :violence-shootself: not a fan.

I've got a phys chem textbook at work I'll have a look at tomorrow but unfortunately it's got the same approach. Maybe my organic chem text might have it.

I do remember something about it, although no lecturer ever referred to it by name, just by how it's used. I think it's pretty much the only thing that stuck about kinetics.

Anyway, what I've guessed about it is that it's simply a way of working out if a reaction is first order (when a collision between two species is the rate determining step... If I remember correctly) by graphing the natural log (ln) of the change in concentration against time. If you get a straight line then it's first order, with the gradient of -k (rate constant).

The way I like to think of it is like a dance. If all the boys and girls at the dance aren't partnered up at the start and then all of a sudden they try to partner up at the same time, they'll form partners really quickly but the ones that miss out straight away find it harder and harder to get a partner until you get the last pair who take a much longer time to find each other. If you graphed the unpaired boys against time you'd find that the graph would be logarithmic, so if you take the natural log instead it should be straight. Chemical reactions that are first order start off really fast but slow down as the reactants have more and more trouble finding each other, just like the dancers.

Do you have any idea about what type of question you'd be expecting on your exam? Maybe you just need to interpret a ln(conc) against time graph and use it to justify whether the reaction is first order or not and maybe also state the rate law after determining k.

Hopefully I can tell you more tomorrow.

[warramungas]

Thanks Db.
It was doing my head in and I've pretty much thrown in the towel as its chewing through too much of my available study time.
As I normally do it mathematically (so long as I have a molarity/concentration and/or a reaction rate), as I dont always have access to an excel spreadsheet, I assume it's to do with kinetics the way you've explained and I understand. If the change in concentration is linear the reactant is zeroeth order which is pretty easy to spot. Then ln[A] 1st order then 1/[A] (not always though) for 2nd order to see if there's any linearity.
The thing he stuffs me up with is he throws in partial orders to throw us out of whack cause they're obviously multistage reactions but not stated how they progress.
I think the only good thing is its supposed to give you the equation of the line and 'k' which is the slope.
Dunno. I'll just keep doing it the hard way.
Cheers.
Warra.

[db1979]

Yeah good approach, don't spend so much time on it that you don't do justice to everything else. I guess I must have been lucky, the bloke that did most of my phys chem lectures followed the KISS principle pretty damn well. I don't remember doing anything on 1/2 orders etc... It was mentioned but we just stuck to the basics. My course seemed to focus a lot on organic and analytical chem, there was very little phys chem that didn't relate to analytical or organic (spectroscopy etc).

Good luck, let us know how you go in your exam.

[Minpac]

Been a good decade and a half since I've done any serious chem/maths. Looks like it's a least squares(non-linear regression curve) approach to working out reaction rates/constants, especially useful if you don't have the start/finish readings or non-regular time based readings.

Check here - useful if you don't have the initial readings from an experiment:

http://andromeda.rutgers.edu/~jimms/kinetics.html

[db1979]

That's the website I looked at. I reckon warramungas has got the gist of it and the lecturer's just being a prick about it, overcomplicating things for kicks.

[warramungas]

Good website minpac. Will probably have a look at it tonight on my laptop as my phone isn't displaying it properly.
Can't understand why we need to gin around nowadays when it so much easier to pull the data for reactions or chemical data from reference tables in chemical si books, which of course you would in real life. I doubt many people mess around working this stuff out freehand anymore.
Cheers guys.
Warra