FuriousGeorge - 8-9-2009 at 16:57
Let's say I have some Fatty Acids, and I want to esterify them with alcohol using H3PO4 as a catalyst. I don't know how much H3PO4 I need, but I do
know:
How many mol of the Fatty Acid I have
How much MeOH I need as a reagent
How much H2SO4 I would need to catalyze it efficiently
Can I derive how much H3PO4 I need?
Ozone - 8-9-2009 at 18:41
Beginnings.
1. Which fatty acid? This is rather important if you would like to know how many moles of "it" you have.
2. See above. You will need an excess of alcohol over the stoichiometric equivalent. Hydrolysis (of the resulting ester) and dehydration (of the
alcohol) will be side reactions which will result in lowered yields.
3. H2SO4 is not H3PO4.
I find that 0.25 mole % works well (of H2SO4 ~ 0.5 % catalyst). Although, I prefer to use a strong acid (sulfonated styrene
VB) resin because I can filter it out an re-use it.
What doe this have to do with acid "strength" and "catalyzation"?
"Strength" has been addressed here, recently, and catalysis depends upon what reaction you wish to do, e.g. esterification or dehydration (and then
which acid to use depends upon what type of hydroxyl you have).
That should answer your questions. Look up the mechanism and work out the stoichiometry.
Cheers,
O3
[Edited on 9-9-2009 by Ozone]
FuriousGeorge - 8-9-2009 at 19:45
I had posted a reply, but I just realized I misunderstood you.
The fatty acids are the ones in vegetable oil (the ones that are free of a glyceride) but I can never be sure of which and in what ratio exactly.
There is probably some animal fat in there too.
I'm fairly certain anhydrous H3PO4 is an appropriate catalyst, based on what I've read.
I just don't know how much to use, to how to calculate that. I suppose titration would work, but I was hoping someone could help me guesstimate.
[Edited on 9-9-2009 by FuriousGeorge]
not_important - 9-9-2009 at 01:51
It isn't absolute "strength" or "weakness" so much as is the acid strong enough to protonate the carboxylic acid, and what reaction rate is acceptable
- the more catalyst the more reactions per second you get (within limits). An acid much stronger than the carboxylic acid will mostly be in the
ionised state, having transferred a proton to the carboxylic acid, while one not much stronger than the carboxylic will not have as high percentage
ionised - meaning a lower percentage of the carboxylic acid gets protonated. More of the weaker acid will compensate and get more of the carboxylic
acid protonated.
For esterfication both H2SO4 and H3PO4 effectively act as monobasic acids as their second ionisation K is low enough to not be very good at
transferring that proton to the carboxylic acid. In either case the mineral acid is used in amounts of 1/100 mole ratio, or less, to the
alcohol-carboxylic acid; a trace of the mineral acid. Stoichiometry really has little to do with it save to give you a rough idea of how much acid is
needed, in many cases in the lab you'd just add a few drops of the mineral acid without much regard to the amount of reactants.
FuriousGeorge - 9-9-2009 at 01:52
You just responded as I completely rewrote my post.
It sounds like you answered my question, though. It seems like you are saying either acid should work in a 1/100 ratio by mol or even less, what
matters is how much longer it will take using a weaker acid.
Did I get it?
Thanks for the response.
[Edited on 9-9-2009 by FuriousGeorge]
Ozone - 9-9-2009 at 05:54
If you check your mechanism, you will see that the H+ is regenerated. In either case, you get 2 at pH 7.02 (2nd ionization of H3PO4) or less. The H+
is thus a catalytic species--it is not consumed and is released during the course of reaction.
Thus, a very small amount, say 1:100 (in my example, 1:200) will be effective. The smallest amount which gives a reasonable rate should be used to
minimize the concommitant rates of various side-reactions.
The H+ is the catalytic species--not the acid source, per se.
Best,
O3
zed - 15-9-2009 at 21:25
The amount and type of catalyst, may be less critical than the act of driving the reaction, by removing water as it is formed.
I suppose you could try to calculate how much, of what type of acid, should work best.
But, Chemistry doesn't quite work that way. With so many subtle and contradictory forces interacting, actual experimentation is required.