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Please forgive me if this is a vague question. I have always wondered how a chemical reaction "knows" where its equilibrium should be. For example, using a basic example of Le Chatelier's principle, for some theoretical reaction $$\ce{A + B <-> C + D}$$ If I were to add extra A and B to the tube, the reaction would proceed to the right until a new equilibrium was established. How do the chemicals "know" that I have added extra A and B?

Am I thinking about this incorrectly? Is it not actually a matter of the chemicals "sensing/knowing" which other species are present, but rather their own individual properties and how likely they are to exist in one form or another?

Related to this quandary, I learned in my metabolism course that NADH and NADPH are used to regulate different reactions because the NAD+/NADH and NADP+/NADPH ratios need to be different, and that merely adding the phosphate allows these separate species to have different equilibria. Can NAD+ really distinguish NADH from NADPH?

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    $\begingroup$ There is no sensing or distinguishing at work here, it is just a matter of statistics and probability. If your equilibrium is $\ce{A + B <=> C + D}$, equilibrium means that both the forward and backward reaction have the same likelihood of occurring in a given time. Now if you chuck in more A and B, it simply makes it more likely that A and B collide to form C and D, than for C and D to collide to form A and B. Ergo, you end up producing more C and D. $\endgroup$ – orthocresol May 8 '17 at 20:47
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    $\begingroup$ This is a good question. We spend a lot of time teaching equilibrium chemistry, where we assume equilibrium is reached before we proceed to the next step. This makes sense because it's lot easier to teach, but it does lead to some odd assumptions about how we must reach the equilibrium. $\endgroup$ – Cort Ammon - Reinstate Monica May 9 '17 at 14:07
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    $\begingroup$ It doesn't. Think of how a chemical reaction actually proceeds deep down - one molecule hits another molecule, and it maybe sticks (A + B -> C). At the same time, random collisions and vibrations may break apart one molecule into two molecules (C -> A + B). The more A and B you have, the likelier it is they will collide in a way that creates C (simply because the favourable collisions are more frequent) and vice versa. When the rate of the two reactions is the same, you get an equilibrium - the reactions still proceed, but in both directions equally. $\endgroup$ – Luaan May 9 '17 at 14:07
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What we got here is a dynamic equilibrium, not a static one. For a static one the molecules would need to know when to stop reacting or when to react in a certain way, in a dynamic equilibrium you don't need to know.

Here's a ELI5 explanation of what's happening:

Let's say you've got a garden with an appletree and hundred of apples are on the ground. You don't want them there so you throw them to your neighbor. he's an old man and doesn't want them either, so he throws them back. You are much faster than him and there are hundreds of apples around you, so you don't even need to move to throw them and you can throw them very fast. He on the other hand has only a few apples laying around and has to move quite far, so he's throwing them back much slower. But what happens after time?

Well, there will be a point where there are more apples on the old mans side than on yours, so you have to run a lot to get to those apples while the old man has to move only some steps to reach a new apple. Eventually it will level out and both of you will throw apples at the same rate. That's a dynamic equilibrium. Because there are more apples on the old mans side he can throw them back as fast as you can throw them to him, even if you are faster.

Now here the chemical explanation:

Reaction rates are dependent on the concentration. High concentration equals fast reaction, low concentration equals slow reaction. At the beginning you got only starting material, it will react fast to the product. The product will react back but there is extremly low concentration so it's slow. But after some time you will reach a point where the concentration of the starting material is so low that the reaction is so slow, that it matches the speed of the back reaction which got faster because you got more and more product. At that time there will equal amounts react in both directions, making it look like the reaction is standing still.

For example: If we got the reaction $\ce{A + B \rightleftharpoons C + D}$ and the reaction rate of the forward reaction is $r_1=k_1[A][B]$ and the backwards rate is $r_2=k_2[C][D]$ then we will reach an equilibrium $r_1=r_2$ and that's the case if $k_1[A][B]=k_2[C][D]$.

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    $\begingroup$ Okay, great analogy. So to extend my question into the dissociation of water, it's not that neutral water somehow knows "oh there isn't enough OH- and H3O+" its just that (at a given temperature) Water has a 1/10^7 probability to be dissociated? $\endgroup$ – Arcadium May 9 '17 at 0:17
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    $\begingroup$ @Arcadium: water dissociation is not a good example because we don't have a clear intuitive idea how it happens. In the example $\ce{A + B \rightleftharpoons C + D}$, the forward rate is proportional to the product of the concentrations of $A$ and $B$ because that is how often they meet. If you add $A$ and $B$ the rate will increase because they meet more often. That will generate more $C$ and $D$. The forward will dominate until the concentration of $C$ and $D$ rises to balance. $\endgroup$ – Ross Millikan May 9 '17 at 3:01
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    $\begingroup$ Yeah, yeah, but what's the chemical analogue of the old guy calling the cops on you, huh? :-) $\endgroup$ – David Richerby May 9 '17 at 10:53
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    $\begingroup$ @DavidRicherby I believe that would be when the old guy's reaction causes your glass reaction vessel to burst all over the floor. $\endgroup$ – Cort Ammon - Reinstate Monica May 9 '17 at 14:05
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    $\begingroup$ @Arcadium Each molecule of $\ce{H2O}$ has some (small, constant) probability at each unit of time to dissociate. However, the rate at which $\ce{HO-}$ and $\ce{H+}$ re-associate are dependent on their concentration (that is, how frequently they bump into each other). For pure water, it turns out the rate of re-association of $\ce{HO-}$ and $\ce{H+}$ matches the (relatively constant) rate of disassociation of $\ce{H2O}$ when $\ce{[HO-][H+]} = 10^{-14}$. (It's a little more complicated than that, as you actually need multiple water molecules to get dissociation, but that's the gist of it.) $\endgroup$ – R.M. May 9 '17 at 18:22

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