Le Chatelier's principle and mashing

Does Le Chateier's principle apply to enzymes? Specifically, does the principle apply the enzymatic reaction during conversion of starches into sugar?

Or, more simply, will enzymes convert more starch into sugar if the sugar concentration is less?

Or, even more simply, will I get better conversion of starch into sugar if I use more water during mashing?

That's not how le chatliers principle works. The concentration of the product has to be low to drive the reaction forward. In this case you are lowering the concentration of starting materials. This should in theory slow down the rate of an enzymatic reaction as the reaction velocity is dependent on substrate concentration.

Le chateliers principle applies only to equilibrium situations. Starches being broken down into sugars in a mash is a one way process. Nothing is putting the sugars back together to make starches.

In a one way reaction, lowering concentration will slow down the reaction.

LeChatelier's principle says that if A + B <------> C + D that increasing either A or B (adding reactants) or decreasing C or D (removing product) will cause the reaction to proceed towards the right and conversely. It says, thus, the same thing as the law of mass action: ([C][D])/([A]) = K
where [A] is the activity (think concentration) of A etc. and K is a constant. It doesn't say anything about how fast (or even if) the reaction will take place. That is the subject of kinematics in which the rates of reactions, which can depend on the concentrations of the reactants and products are described.

But yes, LeChatelier's principle would apply. In

long_sugar + H2O <---> 2 short_sugar

the amount of short_sugar created would increase if more long_sugar were provided or if the produced_short sugar could be removed. But of course it can't. Adding more water would decrease the concentration of short_sugar but it would decrease the concentration of long_sugar as well. As the solutions are fairly dilute the concentration of water wouldn't change that much if you added more and so while technically the reaction would be pushed to the right it wouldn't be by much. That's from the equilibrium point of view. Adding water would also decrease the concentration of the enzymes and that would slow the reaction. Reaction rate might also be slowed by the dilution of the reactants per whichever kinematics model applies but remember that LeChatelier doesn't say anything about rates.

ajdelange said:

But yes, LeChatelier's principle would apply. In

long_sugar + H2O <---> 2 short_sugar

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Is there really a viable pathway by which sugars would recombine in the mash? I've never thought of mashing as an equilibrium process (at least not from a two way reaction perspective), but I accept that I could be wrong.

Here is my take -- you could really ask this 3 different ways:

Does LeChatelier's Principle apply in a mash for
1. pressure?
2. temperature?
3. concentration?

For starters, it's really not an equilibrium reaction since a) enzymes are not consumed in a nice stoichiometric way, they are not a reactant or product. It is better to think of them as a catalyst. Amylase enzymes are like the 2 thugs holding a guy by the arms so loan-shark water can punch him in the gut. And b) it does not move both ways. That's the big one if you're specifically talking about LeChatelier equilibrium. That's not to say there aren't ways to get more complete conversion, but if we're being technical and asking about LeChatelier, it's already out the window.

1. Pressure Since these exist as liquid (water) and aqueous (minerals, starches, sugars, enzymes) states, pressure is out.

2. Temperature Let's make an assumption and see what happens.
Heat + long_sugar + water <--amylase--> 2 short_sugar

There are a few problems here. Amylase does not facilitate recombining longer sugars. If you converted the long_sugars 100% and let the mash sit longer, it does not move backward to find an equilibrium of long_sugars. This also implies adding more heat would continue to drive this to the right. Too much heat would denature the amylase so clearly we won't get more short_sugar by adding heat.
It must be more like:
long_sugar + water --amylase(@heat/pH)--> 2 short_sugar
Since it's not specifically endo- or exothermic (to any appreciably significant number as far as I know), we can't write heat as a product or reactant. Therefor changing heat on either side does not draw the process one way or the other. It's just a requisite for the catalyst.

3. Concentration So let's look at just the reaction now.
long_sugar (aq) ---> 2 short_sugar (aq)
You can add more long_sugars to get more short_sugars, sure. But can you dump in table_sugar and have some of it combine to long_sugars? No. It's not LeChatelier.

My last post didn't make it some how.

Anyway, the hydrolysis of starch is obviously an reaction that goes to equilibrium as most of the starch gets converted in the first 15 minutes and virtually all of it in an hour. And obviously it occurs in the reverse direction as the starch that is being converted to maltose in the mash tun was put together from glucose in the first place. Given that these are reactions that eventually come to equilibrium obviously LeChatelier (or, more precisely, mass action) can be used to explain the extent to which they do go. The problem is that we don't have the whole story. Enzymatically catalyzed reactions can be quite complex and involve several steps each of which is describable by an equilibrium constant (and one or more kinematic constants).

As an example of how such things work in nature consider a reaction I understand better: the conversion of acetaldehyde to ethanol as the last step of fermentation or, in the opposite direction, i.e. metabolism of alcohol in our guts the conversion of Ethanol to acaetaldehyde. The reaction, catalyzed by the enzyme alcohol dehydrogenase is:

CH3CHO + NADH + H+ <---> CH3CH2OH + NAD+

Though it may not at first be obvious the thing which determines whether the reaction goes to the left or right is pH. Thus yeast buffer to a fairly low pH pushing the reaction to the right whereas if we want to assess their work (i.e. determine the amount of alcohol in a solution) we can do that by supplying NAD+ and ADH while buffering to a high pH so that H+ is absorbed and the reaction proceeds to the left (there is a commercial test kit that works this way). Thus it is LeChatelier that explains how fermentation and the test kit work but pH is the thing that drives which way the reaction goes.

I'm not saying that it is this mechanism that is responsible for the hydrolysis of starch or its synthesis as that reaction is quite different from this one but rather that if we knew all the details we could doubtless apply LeChatelier to the hydrolysis reaction. After all, are there any reactions which do not obey the law of mass action?

AJ,

Your point is well taken that Le Chatelier can be applied to any reaction. I guess my point was just that the degree to which the reverse reaction is going to happen in a mash seems so infintesimally low that for practical purposes, it can be considered a one way reaction.

Oh, yes. Definitely.

BrewKnurd said:

AJ,

Your point is well taken that Le Chatelier can be applied to any reaction. I guess my point was just that the degree to which the reverse reaction is going to happen in a mash seems so infintesimally low that for practical purposes, it can be considered a one way reaction.

Click to expand...

As usual he is right on.

I guess I was just implying that when the K value (the equilibrium constant) is soooo small or large (in this case large for the simplified, overall reaction) under mash conditions.... you can essentially "round off" LeChatelier for practical purposes.

There is still some cool chemistry going on, but as a homebrewer it will be very difficult to actually take advantage of the fact that it's in equilibrium.

If that makes sense...

GotPushrods said:

As usual he is right on.

I guess I was just implying that when the K value (the equilibrium constant) is soooo small or large (in this case large for the simplified, overall reaction) under mash conditions.... you can essentially "round off" LeChatelier for practical purposes.

There is still some cool chemistry going on, but as a homebrewer it will be very difficult to actually take advantage of the fact that it's in equilibrium.

If that makes sense...

Click to expand...

And it would appear we are indeed all on the exact same page.

BrewKnurd said:

Le chateliers principle applies only to equilibrium situations. Starches being broken down into sugars in a mash is a one way process. Nothing is putting the sugars back together to make starches.

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Ah, yes, that makes sense. I wondered that it didn't apply.

Equilibrium is a weird word. It has a common definition and technical definition that is the same - forward and reverse reactions at the same rate resulting in no net change (of whatever). However "commonly" people think of this in the realm of whole numbers (i.e. 50/50, even 95/5) yet technically, something can be at equilibrium at ratios like 1 million to 1

pjj2ba said:

Equilibrium is a weird word. It has a common definition and technical definition that is the same - forward and reverse reactions at the same rate resulting in no net change (of whatever). However "commonly" people think of this in the realm of whole numbers (i.e. 50/50, even 95/5) yet technically, something can be at equilibrium at ratios like 1 million to 1

Click to expand...

Absolutely. And even taken to the extreme of a hypothetical reaction that was absolutely one way, with no way for the reverse reaction to occur, it would still meet the defintion of equilibrium once the reaction was complete, as the forward and reverse reactions would both be happening at the same rate, that being not at all.

I guess a formal definition would say that all derivatives of the state variables are 0 meaning that if the system is not disturbed nothing will change no matter how long you wait.