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So, we analyze for the first 12 hours, and then we'll see the results. We can increase the time between analyses to improve accuracy, but then we'll sacrifice the number of tanks we can check.

But what if the poison is in one of the unchecked tanks?!

Is it possible, by experimenting on condemned prisoners, to identify a poisoned tank of kvass, given that even a negligible dose of poison is fatal, and the poison takes effect after 12 hours?

Yes, that's what theory suggests. It's possible to identify it if you're lucky, or it might not be possible. )))) This is the absolutely correct answer to the question. It can be identified, and it might not be.
Anyway, I'm thinking about a solution that would guarantee identification, regardless of luck...

Basically, the task is: within 24 hours, determine which tank is poisoned. Is it possible to do this?

Well, Uranus is right. That is, even if there's only one prisoner, what's the problem? Give him a portion of kvass, then another (from a different barrel) a second later, and so on, a total of 240 portions. After 12 hours, we check: did he die or not? 12 hours and 1 second: did he die or not? 12 hours and 2 seconds – ? And so on. The second he dies is the second the poison took effect.
It all depends on the accuracy of the recording and the accuracy of the "poison takes effect after 12 hours" indicator.

Yes, if death from poison occurs after exactly 12 hours, with an accuracy of +/- less than 2 minutes. We'll track who drinks what and from where, every 4 minutes, within the first 12 hours.

Yes, if death from poison occurs after 12 hours with an accuracy of +/- less than 2 minutes.

But if the accuracy is +/- half an hour... then a different approach is needed.

My apologies, I didn't phrase it accurately. Here's a correction: the poison will kill the condemned person within exactly 12 hours, but the onset of its effects can vary, with a maximum delay of 12 hours, and you need to reliably find the poisoned tank. It doesn't necessarily have to be 12 hours, it could be sooner, but definitely no more than 12 hours. I apologize for providing an incorrect condition that misled you regarding the method of solving the puzzle... :(

Added 9 minutes later
So, is it possible to identify which of the 240 tanks is poisoned?

Yes, definitely it’s possible, otherwise the question wouldn’t have been asked. I think the method involves something similar to bags of fake coins and scales.

Both a "yes" and a "no" answer should be justified.

How can you determine which of the 240 tanks is poisoned?

I'll still nitpick the wording, again: you can. In any case, it can be done randomly. But that's not the point; the question is clear.

How do I do that? What is the exact sequence of actions?

Added 1 minute ago
Well, it should definitely be possible to identify it, if it isn't already implied by default.

Here's my plan.
We give each of the 5 prisoners 240 units of kvass from 48 tanks.
Then, we divide each batch of 48 tanks into 4 groups of 12. And the remaining prisoners from each batch drink, if they haven't already.
In the end, 2 prisoners have drunk kvass from each tank.
We wait 12 hours and identify 2 losers. We have 24 unidentified tanks left, since 2 out of 5 is 10 possible outcomes.
We repeat the process. Here, we divide 24 into 3 groups of 8. And 8 into 2 groups of 4. Accordingly, 2 more prisoners drink from each tank.
The first prisoner drank from tanks 1-8, 9-12, 17-20.
The second prisoner drank from tanks 1-4, 9-16, 21-24.
The third prisoner drank from tanks 5-9, 13-16, 17-24.
In general, we'll have to discard 4, but I probably can't come up with anything better. Or we can experiment with the survivors, if they start dying before 12 hours.

Let's say we only have one round, meaning we need to determine the poisoned barrel within 12 hours, not 24. Then, we have severely limited information from the prisoners. The optimal approach is combinatorics.
We take the first barrel and give its contents to all five prisoners. If the first barrel is poisoned, they will all die within 12 hours.
We take the second barrel and give its contents to all prisoners except the first. If the second barrel is poisoned, the last four prisoners will die within 12 hours. The same applies to the third, fourth, fifth, and sixth barrels.
Then, we combine everything... The last barrel in this list will not be given to anyone, meaning if it is poisoned, everyone will remain alive (hooray!). How many barrels can we test this way? Using the formula:
C5^5 + C4^5 + C3^5 + C2^5 + C1^5 + C0^5, where Cx^y = y!/((y-x)!x!)
= 1 + 5 + 10 + 10 + 5 + 1 = 32
It's a modest result.
Two rounds in this case would be like having ten prisoners? But no... If someone dies in the first round, we can't use them in the second, unfortunately. However... why not take advantage of this and express it as a formula?
So, we take the first barrel and give it to all five... Do we need to repeat this in the second round? No, that's enough; we won't get any more information.
We take the second barrel and give it to four, excluding the first... In the next round, we can give it to the first one or not. We have two options. Accordingly, either four die, excluding the first, in the first round, and then the first dies, or four die, and the first doesn't. Thus, we can get two different results...
In total:
C5^5 * C0^5 + C4^5 * (C0^1 + C1^1) + C3^5 * (C0^2 + C1^2 + C2^2) + C2^5 * (C0^3 + C1^3 + C2^3 + C3^3) + C1^5 * (C0^4 + C1^4 + C2^4 + C3^4 + C4^4) + C0^5 * (C0^5 + C1^5 + C2^5 + C3^5 + C4^5 + C5^5) =
1 * 1 + 5 * 2 + 10 * (1 + 2 + 1) + 10 * (1 + 3 + 3 + 1) + 5 * (1 + 4 + 6 + 4 + 1) + 1 * 32 =
1 + 10 + 40 + 80 + 80 + 32 = 243
In total, we can test up to 243 barrels using these combinations.

Answer: yes

(^ is not an exponent, but a superscript)

We have 24 unidentified tanks left, as 2 out of 5 represents 10 possible outcomes.
We repeat the process. Here, 24 is divided by 3, resulting in 8. And 8 is divided by 2, resulting in 4. Accordingly, 2 units were taken from each tank again.
The first person drank from tanks 1-8, 9-12, and 17-20.
The second person drank from tanks 1-4, 9-16, and 21-24.
The third person drank from tanks 5-9, 13-16, and 17-24.
In total, we'll have to discard 4 tanks, but I can't think of a better solution. Or, we can experiment with the remaining survivors if they don't die within 12 hours.

I didn't understand this. After we do what you suggested earlier, we will have identified 12 tanks, one of which contains the poison (why 24 though??), three living subjects, and one attempt.

Added 2 minutes later
Ment seems to be right. It is indeed possible to determine the poison even in 243 barrels. I took 240 for a round number. I'll check again just in case.

We only have 5 villains and one spare (no one will drink) = 6 piles/indicators.

If time is not a factor, then yes, it's simple.

But with time constraints, you just have to combine them.