[scientizscht (OP)]

Hello

If I have two containers, one with a solution and another with plain water and I connect them with a tube, when will they reach same concentration?

Is there a simple (non differential) equation to calculate that time?

Thanks!

[paul cotter]

Way too many variables missing to even begin to answer such as length and diameter of tube, what is in the tube at t=0, relative heights of containers. Unless the tube length is less than it's diameter I would guess any diffusion to be extremely slow. I would doubt that there is a rigorous equation to describe this.

[alancalverd]

Never.

The solution is in A, the solvent in B

The probability of a solute molecule molecule moving in either direction depends on the relative concentrations {A} and {B}  so the net transport rate  d{A}/dt  per unit time depends {A} - {B} and as {B} -> {A}, so d{A}/dt -> 0.

[Bored chemist]

The probability of a solute molecule molecule moving in either direction depends on the relative concentrations

How would it know?

The probability of a molecule moving in either direction is 50:50.
But net the movement of molecules as a whole is from the high to the low concentration, simply because there are more molecules where the concentration is higher.

If you regard the solution as homogeneous, then the answer is, as Alan says, "never".
But if you are looking at particles then you reach a point where the difference in concentrations is comparable with the natural fluctuations in concentration.

Imagine your solutions were so dilute that you only had 2 molecules (A and B) of solute.
There are 4 possible outcomes;
Molecule A is in container 1 or container 2
Molecule B is in container 1 or container 2

About half the time, both molecules will be in the same container.

So it pretty much starts off at "equilibrium".

[Bored chemist]

Is there a simple (non differential) equation to calculate that time?

Not really.
I presume that's why Alan decided to use a differential one. (Either that or he didn't read what was asked).

But for any given set up of solute, containers and pipes, the rate of mixing is first order decay, with a half life.

[alancalverd]

The probability of a solute molecule moving in either direction depends on the relative concentrations {A} and {B}  so the net transport rate  d{A}/dt  per unit time depends {A} - {B} and as {B} -> {A}, so d{A}/dt -> 0.

Apologies for oversimplifyng the model. Pedantically and etymologically, substitute "the probability P of net migration of molecules from A to B...." which is surely obvious.   

And you can ignore the differential too. Say P_^(A→B) = f({A} - {B}), so as {A} → {B} so P → 0.

[Zer0]

@scientizscht

pi = iMRT ?

[scientizscht (OP)]

@scientizscht

pi = iMRT ?

Thanks but where is time? 🤔

[Zer0]

@scientizscht

pi = iMRT ?

Thanks but where is time? 🤔

Won't Time be a Dependent Variable?

ps - an easy way to know " t " would be to conduct the Experiment.

" Answer thru Demonstration shall tc of that. "
(Isaac)

[Zer0]

@scientizscht

On second thought...
Your point seems Valid.

There should be a way to incorporate  ' t ' into any Equations/Formulas.

Even if it's just a Symbolic representation, it would still define the Interdependence.

[hamdani yusuf]

pi = iMRT

IMO, it's hidden in the dimension of osmotic pressure pi, which is ML(-1)T(-2).
The higher the value of pi, the lower the equilibrium time will be. It can be done by increasing temperature.
As you said, diameter and length of the connecting tube will also be influential.

[Bored chemist]

pi = iMRT

IMO, it's hidden in the dimension of osmotic pressure pi, which is ML(-1)T(-2).
The higher the value of pi, the lower the equilibrium time will be. It can be done by increasing temperature.
As you said, diameter and length of the connecting tube will also be influential.

Thanks for clarifying that pi means the osmotic pressure rather than pi.
That makes  more sense now.

But rates of reaction are not strongly related to osmotic pressure.
Osmotic pressure is, itself, an equilibrium process and "getting to equilibrium" has kinetics of its own and an associated timescale.

[hamdani yusuf]

But rates of reaction are not strongly related to osmotic pressure.
Osmotic pressure is, itself, an equilibrium process and "getting to equilibrium" has kinetics of its own and an associated timescale.

Imagine an extreme case where the solution is nearly identical to the solvent. There will be nearly zero osmotic pressure when the containers are connected, and there will be low flow of solvent to the container of the solution.
Hydrostatic pressure will reach equilibrium much earlier than equilibrium of concentration.
Osmotic pressure depends on concentration difference, which will be reduced when material flow is allowed. And you already mentioned about decaying process.

[Bored chemist]

A big molecule will have the same effect on osmotic pressure as a small one, but it will diffuse more slowly.

[hamdani yusuf]

I don't think that osmotic pressure describes a complete picture to the situation asked in this thread. But it's close enough to be a starting point towards the real solution.
The molecules should move from one container to the other because of some kind of force. Let's just call it chemistromotive force, as an analogy to electromotive force coined by Faraday.
As mentioned before, that force depends on several factors, such as temperature, sizes of the connector, difference in concentration in each type of molecules or ions. Not just the difference in total osmotic pressure between the containers.
If one container contains acid, while the other contains base, then the motive will be higher than if they both contain acids or bases, or neutral.
Even if the opposite containers contain very similar molecules, like enantiomers, there will be a motive toward equilibrium, although the osmotic pressure is close to zero.

[Bored chemist]

The reason why molecules diffuse is simply "because they can".
The only driving "force" is entropy. (which is more or less why they all work equally well to raise osmotic pressure.)

If one container contains acid, while the other contains base, then the motive will be higher than if they both contain acids or bases, or neutral.

How will the molecules in one place know what molecules are in another place, in order to move towards it?

[hamdani yusuf]

How will the molecules in one place know what molecules are in another place, in order to move towards it?

I'm not sure yet, but perhaps it's related to electromagnetic fields. Suppose one container contains Na+ and OH- ions, while the other contains H+ and CL- ions. H+ ions are the lightest. For them to have the same temperature as the others, they must move faster in average. They would move to the other container quicker. Electrostatic charges imbalance then forces the movement of the other ions to rebalance them.

[Bored chemist]

I'm not sure yet, but perhaps it's related to electromagnetic fields.

We know what happens in the presence of such fields
https://en.wikipedia.org/wiki/Debye%E2%80%93H%C3%BCckel_theory

But no such field exists.

My question was rhetorical.
The molecules can not possibly know what is happening somewhere else.
So there's no way for that to affect the way they diffuse.

(It might stop them diffusing back- which would affect the net transfer rate)

[hamdani yusuf]

My question was rhetorical.
The molecules can not possibly know what is happening somewhere else.
So there's no way for that to affect the way they diffuse.

Do you think H+ diffuse at the same rate as Na+ at the same temperature?

Suppose one container contains Na+ and OH- ions, while the other contains H+ and CL- ions. H+ ions are the lightest. For them to have the same temperature as the others, they must move faster in average. They would move to the other container quicker. Electrostatic charges imbalance then forces the movement of the other ions to rebalance them.

[Bored chemist]

Do you think H+ diffuse at the same rate as Na+ at the same temperature

As I said

A big molecule will have the same effect on osmotic pressure as a small one, but it will diffuse more slowly.

Please try to keep up.