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A layman's guide to Le Chateleir's Principle

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  • A layman's guide to Le Chateleir's Principle

    The foundation of my Foolproof Room Temperature biodiesel method is Le Chateleir's principle. For me, as a chemist, it's a very simple and straightforward principle, but based on some of the questions and comments I have received, it's not so straightforward for others.

    Fortunately, it's a proniple that has applications in many areas of life, so I'll have a go at putting it in laymans terms.

    In broad terms, chemical reactions can be divided into those that react completely, and those that form an equilibrium:

    1. Complete reaction: A + B -> C + D

    An example is the reaction of hydrochloric acid with caustic soda to form salt and water: NaOH + HCl -> NaCl + H2O

    That is, if the acid and base are added in a 1:1 ratio, they are all consumed, and all that is left after the reaction is salt and water

    2. Dynamic equilibrium: A + B <-> C + D

    During this process, we initially add A to B. They begin reacting to form C and D. This is the forward reaction. But then, as the concentration of C and D begin to increase, they begin reacting with each other to reform A and B. This is called the reverse reaction. Eventually, the rate of the reverse reaction is the same as the forward reaction. When this point is reached, it is said to be a "dynamic equilibrium." Under these circumstances, the concentrations of all species have stabilised and are no longer changing. This happens despite the fact that both the forward and reverse reactions are still proceeding. But because both reactions are proceeding at exactly the same rate, there is no net change.

    This equilibrium can be represented mathematically. The general formula for this type of thing is reactants/products, so it would be:

    K = [C].[D]/[A].[B] where

    K = equilibrium constant

    [X] = concentration of X

    The reason that this equilibrium can be represented by a constant, is that when these 4 species are present in solution, their mathematical relationship will always be determined according to this equation. If the concentration of A-D are changed by adding or removing components, the concentrations of all 4 components will change to ensure this mathematical relationship is preserved.

    So for example, suppose that at equilibrium the concentration of all components was 1M. The Equilibrium Constant would be 1x1/1x1 = 1

    Now suppose that I increased the concentration of A and B to 3M. This will cause the rate of the forward reaction to increase and the reverse reaction to decrease. 1 equivalent of A would react with 1 equivalent of B to form 1 equivalent of both C and D. So now A and B will have gone from 3M to 2M, and C and D will have gone from 1M to 2M. Thus, equilibrium has now been reestablished, and Keq = 2x2/2x2 = 1

    Let's illustrate this with an analogy. This principle, also known as a negative feedback loop, has many applications in the world around us.

    Suppose you are attending a function at a function centre. There is an inside area and an outside area. As you watch you notice some people are walking from inside to outside (forward reaction) and others walking from outside to inside (reverse reaction).

    You snap a photo of the interior area and then go outside and snap a pic of the outside area. You count the people and discover that there are 50 outside and 50 inside. So you come up with an equilibrium constant K = outside/inside of 50/50 = 1.

    Now suppose a coach bus arrives and 50 new guests enter the building. Suddenly there are now 100 inside and 50 outside. It's pretty easy to see that it would now seem rather crowded inside compared to outside, so there'd be a net flow of people from the inside to the outside.

    In terms of the equilibrium, this means that the forward reaction is now proceeding at a greater rate than the reverse reaction.

    Eventually, the numbers would probably stabilise, probably with about equal numbers inside as outside. When it reaches this point, we have again achieved a dynamic equilibrium, with the rate of the forward and backwards reactions being equal, with people moving from the inside to the outside at about the same rate as those moving from the outside to the inside.

    Thus, we again have an equilibrium constant of 1 (Keq = 75/75).

    Suppose that now an ice-cream truck pulls up outside, and 20 people leave the area to go across the road to buy an ice cream, and then standing around talking.

    Now the numbers outside have gone from 75 to 55 outside, and since there are 75 inside, outside will now seem a little less crowded, and it's pretty easy to see a slight shift in the equilibrium with 10 people drifting outside to make it 65/65.

    This is obviously only a contrived an example but it illustrates the concept of a dynamic equilibrium, and how it can shift according to the experimental circumstances.

    Let's now consider this in the context of biodiesel manufacture.

    Dr Pepper Method:

    KOH or NaOH are added to MeOH. This sets up the following dynamic equilibrium:

    KOH + MeOH <-> MeOK + H2O

    That is, at equilibrium, the potassium hydroxide, the methanol, the methoxide, and water are all present. And for every molecule of methoxide generated, one molecule of water is generated. Most people that use this method seem to be unaware of this. There are four potential sources of water in this mixture:
    • water in the oil
    • water in the methanol
    • water in the KOH
    • water from the methoxide reaction.

    A source of great puzzlement to me is the effort that people go to in order to ensure dry oil, while they are unaware that they are adding water to the mixture with their methoxide solution. I can tell you that of all the sources of water, water in the oil is the least significant.

    I don't know what the equilibrium constant is with this reaction, but all four species are certainly present.

    So when it's added to the WVO, at least 4 reactions occur:

    1. Neutralisation of FFA with KOH: RCOOH + KOH -> RCOOK (soap) + H2O
    2. Neutralisation of FFA with methoxide: RCOOH + MeOK -> RCOOK (soap)+ MeOH
    3. Transesterification of triglyceride to form biodiesel and glycerol: C6H5O6R3 + 3MeOH -> 3(MeO2CR) + C3H8O3
    4. Saponification of triglyceride: C6H5O6R3 + 3KOH -> 3(KO2CR) + C3H6O3

    Let's now consider the role that the saponification reaction plays in effecting the equilibrium.

    In the transesterification reaction, the KOH is a catalyst. That's why it doesn't appear in the stoichiometric reaction. In other words, it plays a part in the reaction mechanism, but is not consumed, so it is regenerated afterwards. But in the saponification reaction it is not a catalyst but a reactant.

    In other words, the saponification reaction consumes the KOH. This will pull the equilibrium to the left: KOH + MeOH <- MeOK + H2O

    This is probably the principal reason that the Dr Pepper method requires heat. Whatever KOH is left over from the neutralisation of the FFAs, it is ameliorated by the inevitable saponification reaction. And you can't overdose the mixture with KOH because it will
    • make even more soap
    • make even more water

    The combination of these two factors may, of course, mean that your batch emulsifies, and you finish up with white gloop instead of biodiesel, and there are many instances of this happening on this and other forums. This is the great weakness of the Dr Pepper method - to make biodiesel you must navigate between the interfering reactions.

    The FRT method, however, eliminates these interfering reactions, and you can add as much KOH as you want.

    FRT Method:

    KOH or NaOH are added to MeOH. This sets up the following dynamic equilibrium:

    KOH + MeOH <-> MeOK + H2O

    The water is unwanted component of this process, so I simply eliminate it by adding a drying agent.

    Consider now what happens to the equilibrium if water is removed. According to Le Chateleir's principle, the reaction will now react to the right, to attempt to generate more water:

    KOH + MeOH -> MeOK + H2O

    But if the water is removed as soon as it is formed, then the reaction will continue to react until all the KOH is consumed. So after this process has gone to completion, we now have only MeOK and MeOH in the solution. It therefore eliminates reactions 1 and 4 above, so there are now only two reactions:

    1. Neutralisation of FFA with KOH: RCOOH + KOH -> RCOOK (soap) + H2O (eliminated)
    2. Neutralisation of FFA with methoxide: RCOOH + MeOK -> RCOOK (soap)+ MeOH
    3. Transesterification of triglyceride to form biodiesel and glycerol: C6H5O6R3 + 3MeOH -> 3(MeO2CR) + C3H8O3
    4. Saponification of triglyceride: C6H5O6R3 + 3KOH -> 3(KO2CR) + C3H6O3 (eliminated)

    As reaction 4 has been eliminated, all the MeOK is now available for catalysis of the transesterification process. This allows the procedure to go to completion at room temperature.

    And also, as there are no interfering reactions, you can add as much KOH to the mixture as you want. In ten years or more of using this method, I've used a 10% KOH solution at a 100:15 ratio of WVO:MeOK and never had any problems. The ratios have now changed, however, to 100:20 sor there is even more metoxide present. If I encontered a WVO that was particularly high in FFAs, for example, I can just increase the amount of KOH with no interfering processes occuring.

    Ok - that's it. As before I'm happy to answer any genuine questions, but the posts and responses from a certain individual will be ignored.