# Why doesn't capture of infinitely rare ions by microbes in an infinitely large ocean cost infinite free energy?

Sorry if the question title is misleading, but:

Suppose we have a single "cell" with internal volume $$V_c$$ and with a single potassium receptor, in an ocean of volume $$V$$ which contains a single potassium ion.

Assume that the receptor, when the ion is near enough, has a 100% chance of capturing and putting it inside the cell.

My confusion is as follows:

• If the ion is at first in the ocean, at any unknown point (imagine an ensemble which has its position uniformly distributed) then by capturing the ion, the cell seems to reduce the entropy of the ion by $$log(V/V_c)$$. Therefore if we imagine hypothetical very very large oceans, it should cost free energy proportional to $$log(V)$$ to capture the ion? if we imagine an unrealistically huge ocean it would cost an obsene amount of energy.

• Yet in actually existing cells, it seems like it typically costs a fixed amount of energy (I think the content of one ATP molecule) to capture an ion like that. so therefore the previous argument must be wrong.

This is a contradiction: On the one hand, my theoretical argument suggests that the free energy cost depends on the volume of the ocean. Yet real world receptors cost a fixed amount of energy to use once. This implies that we can in theory set the volume of the ocean large enough so that the free energy cost is larger than the energy cost of a onetime use of the receptor.

Clearly this is a contradiction, so where do I misunderstand the issue?

EDIT: I am not sure this helps, but because of some comments: we could imagine that the receptor is automatically removed from the membrane after the ion is captured, so that the ion cannot travel back.

EDIT: There was some criticism of the 100% assumption. I just made that for simplicity. I am still equally unsure what the answer would be if we assumed 90% chance of capturing it.

• statistical mechanics sometimes does weird things when you use it for single particles, because it's meant to describe the behaviour of a large number of particles without having to account what every single particle is doing on its own. Commented Jul 22 at 20:27
• @paulina, I'm not sure I buy this... Liouville's theorem is exact for any number of particles isn't it? besides, you could just increase N and V in the right amount to balance each other out in terms of entropy decrease and we get the same issue. Though od course I must be mistaken somewhere. Commented Jul 22 at 20:33
• Whether something does or does not happen doesn't tell us anything about how much energy the process takes. I don't understand your question in the first place. Commented Jul 22 at 22:42
• @FlatterMann, I know it doesn't but I'm not sure what that has to do with my question. Does my new text under "this is a contradiction" clarify? Commented Jul 23 at 3:45
• Honestly the habit in this stackexchange community of just closing questions with largely no explanation is just stupid. You could have just told me what is unclear about the question and I would have tried to clarify it. Commented Jul 23 at 4:39

If a K+ ion has a (near-)100% chance of being captured, then the resulting bond is quite strong, and the corresponding energy release from forming that bond is notable. This energy heats the cell and surroundings and increases the total entropy by more than enough to counteract the positional entropy decrease from immobilizing the ion.

(The same argument, in reverse, justifies why all materials have a nonzero vapor pressure. No bond has a 100% chance of staying intact because thermal fluctuations at any nonzero temperature—which is any temperature—could always arise to break a surface bond and allow an atom/molecule to freely explore the rest of the universe in gaseous form. This provides a notable entropy benefit. Taken together, the two arguments clarify Nature’s preference for both strong bonding and many possible configurations, as mediated by the temperature—reflecting minimization of free energy $$G\equiv H-TS$$, where a low enthalpy $$H$$ signals strong bonding.)

• I don't think this answers the question. My question argues that the entropy increase due to putting the atom from $V$ into $V_c$ is logarithmic in V, but the free energy use of using the receptor is constant. Therefore there is some value of $V$ for which the entropy decrease dominates the energy spent by the receptor, thereby seemingly violating physical law. The fact that the bond is "quite strong" and the energy release is "notable" doesn't matter for that argument. Commented Jul 23 at 4:37
• "Therefore there is some value of V for which the entropy decrease dominates the energy spent by the receptor, thereby seemingly violating physical law." You're assuming that a potassium ion somewhere in the cosmos must necessarily attach to a receptor. Why? As you note, "There was some criticism of the 100% assumption. I just made that for simplicity." I think the assumption violates physical law. Commented Jul 23 at 6:17
• now you're saying a 100% effective receptor assumption violates physical law? earlier you just said that it takes a "notable" amount of energy? which is it? Commented Jul 23 at 6:39
• "ou're assuming that a potassium ion somewhere in the cosmos must necessarily attach to a receptor. Why?" I mean this is basically a kind of ergodicity assumption.if the ion is somewhere in the massive ocean, and it just freely floats, it will "eventually" reach the volume of space within which the receptor is active. Commented Jul 23 at 6:40
• (Ofcourse for my argument the ocean might have to be big enough that it collapses into a black hole, I don't know that because I haven't quantified it, but at this point that doesn't seem essential to me, since I THINK Liouville's theorem also holds with arbitrarily weaker gravitational constants.) Commented Jul 23 at 6:42

If a K+ ion has a $$100$$% chance of passing through a receptor when it gets near enough to the outside, it would also have a $$100$$% chance of passing through when it got near enough the inside.

If the concentration is larger on the outside, ions on the outside will get near the receptor more often. The concentration will increase until inside and outside have the same concentration.

If you want a larger concentration on the inside, you cannot let them pass freely. You have to spend energy to put them on the inside.

• When the concentrations inside and outside are equal, where exactly is that one ion? Commented Jul 22 at 23:37
• @WillO - Wandering around the ocean, spending on the average the same amount of time inside the microbe as in any other volume of the same size. Commented Jul 23 at 1:27
• "If you want a larger concentration on the inside, you cannot let them pass freely. You have to spend energy to put them on the inside." Have you read my question? I'm not sure if you have, given that I say that the free energy cost of a receptor is that of one ATP molecule. Commented Jul 23 at 3:36
• I am afraid the answer doesn't make sense because I completely misread what you are asking. Oops. Commented Jul 23 at 3:36
• @mmesser314, that's ok! I mean the question is closed now by some anonymous person who gave me no indication what was wrong with the question so no way to fix it, so let's just let it go. It's kind of a shame for me since this seems like an important misunderstanding I have. Commented Jul 23 at 4:46