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Thread: The chemistry of the glycerol (and soap) phase

  1. #11
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    Re: The chemistry of the glycerol (and soap) phase

    [QUOTE=Tony From West Oz;66558]
    Here is the equation: KOH + MeOH <-> MeOK + H2O[/]quote
    As you have stated previously,
    Le Chateleir's principle applies, but you do not give the proportions at STP for this reaction.

    Tilly's statement could be true if the reaction is weighted in favour of the solution over the reaction products.
    So here are my questions:
    1. What is the formula for calculating the ratios?
    2,
    So, if I have 1Kg of KOH and 20 litres of 'MeOH', How much 'MeOK' and H2o would be produced and how much KOH dissolved in H20 would be present at any one point in time?

    As a Chemist you should be able to specify the ratios of each side of the reaction at STP.
    I await your reply with the answers to my questions.

    Tony,

    With the greatest of respect, you just simply do not understand this. I've tried, again and again and again, to explain this. I've explained it in great detail on my website, and I specifically constructed a post on this, where I slowly and painstakingly went through through the chemistry of both the Dr Pepper method and my method. I explained the nature of the equilibrium in fine detail, and explained all the various reactions.

    And at one stage I wasn't going to bother even doing this. When I first posted my method a few years ago, I didn't bother answering any of the questions, as I figured I was pretty much wasting my time explaining chemistry to people with no formal qualifications, and many of whom plainly hadn't even done chemistry at high school.

    But I changed my mind, and have now gone into great detail attempting to explain this, now with multiple posts, specifically. designed to be understood by the layman. But I'm now starting to wonder whether I've just wasted my time.

    There is just no more simpler language I can use to explain this. And this should not really be surprising. This is why we have universities, as some things cannot be understood in five minutes by reading a blog post. Le Chateleir's principle is a 100 level concept, but the mechanism and kinetics of other parts of the process are 200 and 300 level concepts. If you don't understand them, it doesn't mean you're dumb, but simply that you don't have the right educational background.

    I once had someone try to explain futures trading to me. I didn't understand a word of it. Concepts that were simple to him were just entirely foreign to me. But one thing I didn't do was to conclude that this guy was wrong simply because I hadn't understood his explanation.

    And so if you still don't understand my explanation, then those on this forum have to decide whose advice they want to take - someone with a PhD in Chemistry, and 30y experience in the chemical sector, and who when communicating on this forum always speaks politely and respectfully with people - or someone with no formal qualifications in Chemistry, who takes great personal offence when anyone disagrees with him, and bombards their threads and posts with loud shouty bold text, and has made this forum such an unpleasant place that many people are now frightened to post.

    You choose which one has the most credibility.

    In any case, I will not be either reading or responding to Tilly's posts. Of all the people I reply to on this forum, he is the one with the least ability to understand simple chemical concepts, and the one with the most childish attitude. It's a complete waste of time to attempt to communicate with him as an adult.

    Having said all that, I'll try again. I've explained it again and again and again and again. But I'll try one last time. But this is my last attempt. After this I give up. I'll try to do it in point form.


    • I don't know the equilibrium constant, and I don't have to. The reason simply is that it is a dynamic equilibrium. That is, it is constantly reacting in both directions, and will shift according to other reactions occurring. No matter how many times I've attempted to explained this, you just still don't understand how a dynamic equilibrium works.



    • Each molecule of KOH reacts with one molecule of MeOH to form one molecule of potassium methoxide and one molecule of water. For every molecule of MeOK you make, you make one molecule of water.
    • With the Dr Pepper method, if your concentration of MeOK isn't high enough, there isn't enough catalyst to carry out the reaction. On the other hand, if it's too high, then you have too much water, and the competing saponification reaction makes enough soap to emulsify your batch and it turns to gloop. That's why you have to titrate - to get a successful reaction you have to hit a pretty narrow window of concentration
    • Because of this, you can't get a high enough concentration of catalyst to permit the reaction to proceed at room temperature.


    This is a very common conundrum in organic synthesis. It's very common to find yourself with competing reactions occurring simultaneously. And so any chemist worth his salt will just simply modify the reaction to favour the desired reaction.

    When I looked at this procedure I saw an obvious modification to me. Since I have now had such immense difficulty explaining this to all you non-chemists, I can now see why it wasn't obvious to others.

    My modification was simply to remove the water. This simultaneously achieves the following


    • it pulls the equilibrium to the right, thus increasing the concentration of catalyst, allowing the reaction to proceed at RT
    • It eliminates the competing saponification reaction, so you can shovel in as much KOH as you want
    • It results in a deprotonated form of glycerol, resulting in lower viscosity and easier disposal.


    Okay that's it. I just can't explain it any more than this and I can't simplify it any more than this. I've explained this over and over and over and over, and if you still don't understand it, enrol in an online chemistry course (if such things exist) or at a university.

    Either that or just turn up and watch me make my next batch. You don't have to understand the chemistry, just follow the instructions;.
    Last edited by Dr Mark; 22nd April 2019 at 09:56 PM.

  2. #12
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    Re: The chemistry of the glycerol (and soap) phase

    Mark,
    I did a quick Google and found (amongst others) http://www.arpnjournals.org/jeas/res..._0816_4863.pdf
    This gives the Molar ratios and the reaction coefficients for different reaction temperatures.
    Now, this should give us the ability to predict the amount of water for the given ratios of KOH and MeOH.
    I believe that Molar ratios are based on the relative mass of the feedstock eg KOH should have K (40), O (16), H (1) = 57, Methanol CH3OH would have C (12) H *4 (4) O (16) = 32
    In this case a molar ratio of 0.012 ( KOH:MeOH) would give mass ratios of (32*0.012):57 or 0.0067. Thus for a molar ratio of 0.012 KOH:MeOH you would need to add 6.7g of KOH per Kg of Methanol.

    For a reaction temperature of 25 and a molar ratio of 0.012, this would give us:
    KOH in solution 1.09%
    CH3OH in solution 96.91%
    CH3OK in solution 1.64%
    H2O in solution 0.81%

    Are my figures accurate?

    Please comment on this as I have only lower school Chemistry.
    Life is a journey, with problems to solve, lessons to learn, but most of all, experiences to enjoy.

    Current Vehicles in stable:
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    '80 Mercedes 300D. 2 tank conversion [Sold]
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  3. #13
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    Re: The chemistry of the glycerol (and soap) phase

    Quote Originally Posted by tillyfromparadise View Post
    Hi Mark,
    Wrong again, as usual.
    You do seem to have a problem understanding what is written.


    H2O is only produced when you actually produce MeOK.
    As you continually point out, when enough water is produced, the Production of MeOK stops. You may have forgotten, but this is an equilibrium reaction. At the point where the production of MeOK stops because of the presence of too much H2O, the production of H2O also stops. No more MeOK or H2O/ water is produced.
    The remaining unreacted KOH is just dissolved in the methanol without producing H2O or MeOK.
    A real chemist going by the name Neutral who had over 30 years R&D in the alternate fuel field including actually making "biodiesel" in the 70's {before it was called biodiesel) pointed that out to me years ago
    .
    At the same time he told me that he thought that not much MeOK was actually produced when you mixed KOH with methanol so not much H2O was produced. He thought it was mostly just KOH dissolved in methanol

    PS. THIS IS SHOUTING. This is emphasizing an important point.
    If you have any further confusion about any of this please do not hesitate to ask.
    I give up. I can see now that it was a mistake to attempt to engage with you at all. As has been the case in the past, as you have no formal qualifications in chemistry, and have deluded yourself into thinking that you understand some chemistry just because you have learned a few big words, your posts are an utterly baffling collection of random chemical terms and concepts that make no sense to anyone but you. Consequently I'm just wasting my time, and will just go back to ignoring everything you say.

    Please refer to my conversations with Tony - unlike you he is an adult, and is keen to learn from someone that knows a lot more about Chemistry than him.
    Last edited by Dr Mark; 22nd April 2019 at 11:39 PM.

  4. #14
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    Re: The chemistry of the glycerol (and soap) phase

    Quote Originally Posted by Tony From West Oz View Post
    Mark,
    I did a quick Google and found (amongst others) http://www.arpnjournals.org/jeas/res..._0816_4863.pdf
    This gives the Molar ratios and the reaction coefficients for different reaction temperatures.
    Now, this should give us the ability to predict the amount of water for the given ratios of KOH and MeOH.
    I believe that Molar ratios are based on the relative mass of the feedstock eg KOH should have K (40), O (16), H (1) = 57, Methanol CH3OH would have C (12) H *4 (4) O (16) = 32
    In this case a molar ratio of 0.012 ( KOH:MeOH) would give mass ratios of (32*0.012):57 or 0.0067. Thus for a molar ratio of 0.012 KOH:MeOH you would need to add 6.7g of KOH per Kg of Methanol.

    For a reaction temperature of 25 and a molar ratio of 0.012, this would give us:
    KOH in solution 1.09%
    CH3OH in solution 96.91%
    CH3OK in solution 1.64%
    H2O in solution 0.81%

    Are my figures accurate?

    Please comment on this as I have only lower school Chemistry.
    Thanks for the paper - I'll have a look at it when I get a chance
    T

  5. #15
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    Re: The chemistry of the glycerol (and soap) phase

    Hi Mark,

    Quote Originally Posted by Mark View Post
    In any case, I will not be either reading or responding to Tilly's posts

    Why don't you grow up Tilly? Seriously, why don't you stop being so childish?
    I've got two degrees in Chemistry and you don't have any. I have tried and tried and tried to explain Le Chateleir's principle to you, particularly as it pertains to this method, and this sort of childish drivel is the best you can do?

    There are people on this thread that want to learn some chemistry. You're not one of them.

    We all know the truth. Your fragile ego is threatened by the fact that someone might have developed a superior method than yours. Welcome to the real world son. Get back to me when you either

    a. Get a degree in Chemistry

    b. Grow up
    I think your vow that you "will not be either reading or responding to Tilly's posts" lasted for an even shorter time than my New years commitment to go on a diet.
    I forget, please remind me how many degrees you have in chemistry

  6. #16
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    Re: The chemistry of the glycerol (and soap) phase

    Dr. Mark; The constant of acidity of methanol is given as 2.9 times ten to the minus 16 power. That's a pH measurement. Methanol is a weak acid as is ethanol. That's the equilibrium constant at standard temperature and pressure for methanol. That's the acid disassociation constant. The biggest problem I've run into is when a quantity of potassium hydroxide is dissolved in methanol the constant of basicity isn't in water as a dilute solution. There's about 10-15% water in solid potassium hydroxide. Then the formation of water when methoxide is produced, escorted by potassium ion. The pKa of methanol is 15.54, but that's in water. It seems to me that since a small amount of water is present the normal equation for calculating eqilibrium of bases (methoxide) doesn't work. I don't know how to calculate the amount of methoxide present in anhydrous methanol with potassium hydroxide dissolved in it. Then adding calcium oxide to the mix which absorbs or adsorbs most of the water present furhter complicates figuring the quantity of methoxide present. Remember using calcium oxide to dry ethanol only produces 99.5% dried ethanol it's not 100%.
    Last edited by WesleyB; 23rd April 2019 at 02:18 PM.

  7. #17
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    Re: The chemistry of the glycerol (and soap) phase

    Hi Mark,
    Quote Originally Posted by Mark View Post
    I give up. I can see now that it was a mistake to attempt to engage with you at all...and will just go back to ignoring everything you say.
    Do you mean you will no longer tell me what an intelligent and well educated chap you are or that you think I am childish and do not understand the Chemistry. That is a great disappointment.
    Last edited by tillyfromparadise; 23rd April 2019 at 04:49 PM.

  8. #18
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    Re: The chemistry of the glycerol (and soap) phase

    Quote Originally Posted by Tony From West Oz View Post
    Mark,
    I did a quick Google and found (amongst others) http://www.arpnjournals.org/jeas/res..._0816_4863.pdf
    This gives the Molar ratios and the reaction coefficients for different reaction temperatures.
    Now, this should give us the ability to predict the amount of water for the given ratios of KOH and MeOH.
    I believe that Molar ratios are based on the relative mass of the feedstock eg KOH should have K (40), O (16), H (1) = 57, Methanol CH3OH would have C (12) H *4 (4) O (16) = 32
    In this case a molar ratio of 0.012 ( KOH:MeOH) would give mass ratios of (32*0.012):57 or 0.0067. Thus for a molar ratio of 0.012 KOH:MeOH you would need to add 6.7g of KOH per Kg of Methanol.

    For a reaction temperature of 25 and a molar ratio of 0.012, this would give us:
    KOH in solution 1.09%
    CH3OH in solution 96.91%
    CH3OK in solution 1.64%
    H2O in solution 0.81%

    Are my figures accurate?

    Please comment on this as I have only lower school Chemistry.

    Right, let's have a look at this. I'm not sure where you got those numbers from, but we can work through this one step at a time.

    As it happens it proves everything I've been saying, but it's nice to have some actual numbers to work with.

    I've attempted to insert some images from the paper, however, which are cut and pastes of the relevant charts and figures, but when I use the Insert Image function it says it is an invalid file, despite the fact that I've tried both jpg and png formats. What am I missing?

  9. #19
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    Re: The chemistry of the glycerol (and soap) phase

    Quote Originally Posted by WesleyB View Post
    Dr. Mark; The constant of acidity of methanol is given as 2.9 times ten to the minus 16 power. That's a pH measurement. Methanol is a weak acid as is ethanol. That's the equilibrium constant at standard temperature and pressure for methanol. That's the acid disassociation constant. The biggest problem I've run into is when a quantity of potassium hydroxide is dissolved in methanol the constant of basicity isn't in water as a dilute solution. There's about 10-15% water in solid potassium hydroxide. Then the formation of water when methoxide is produced, escorted by potassium ion. The pKa of methanol is 15.54, but that's in water. It seems to me that since a small amount of water is present the normal equation for calculating eqilibrium of bases (methoxide) doesn't work. I don't know how to calculate the amount of methoxide present in anhydrous methanol with potassium hydroxide dissolved in it. Then adding calcium oxide to the mix which absorbs or adsorbs most of the water present furhter complicates figuring the quantity of methoxide present. Remember using calcium oxide to dry ethanol only produces 99.5% dried ethanol it's not 100%.
    The paper that Tony sent me contains some very useful data. I'll give a fuller explanation, now that I have some real numbers, as soon as I can figure out how to cut and paste figures from the paper. I saved them as jpg screen captures but for some reason they won't upload

    Wasn't aware of the 99.5% figure for ethanolic dehydration. Where did you get that from? I know for a fact that when organic chemists want to dehydrate an alcohol of any sort they use CaO - nothing works better. But 99.5% is probably good enough anyway. Although I now use cement, not CaO, which has other anhydrous compounds in there.

    Trying to make theoretical predictions in situations like this based on pKas is fraught with danger, for exactly the reasons you cite - different solvents, hydrogen bonding etc.
    T
    Last edited by Dr Mark; 23rd April 2019 at 10:37 PM.

  10. #20
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    Re: The chemistry of the glycerol (and soap) phase

    My reference to drying ethanol is from "A Text-Book of Practical Organic Chemistry" author Arthur I. Vogel , publisher Longman Group Limited , Third Edition, pages 166-167 in the article titled Absolute Ethyl Alcohol. "The product is generally termed "absolute" ethyl alcohol, although it usually still contains about 0.5 per cent of water." In the book the next proceedure is for "super-dry" ethyl alcohol where magnesium metal and iodine then anhydrous distillation is used to further dry grain alcohol to 200 proof. It seems to me that calcium oxide with slighty wet alcohol still has an equilibrium where the ethanol is not perfectly dried.

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