If Brownian motion is driven by random walks, then why does it occur along a concentration gradient?
In: Chemistry
It is a matter of probability.
Imagine you have two compartments. Compartment A contains only 1 particle, whereas compartment B contains 100. If they all drift about randomly, it is simply more likely that particles will cross from B to A, since there are so many more in B than in A. This will continue until there are 50 particles in both A and B. At this point the rate at which they cross over is identical, as there are an equal amount of particles in both compartments.
There are exceptions to this, as when charge is involved or the organism exerts energy to determine the direction of the flow.
This is how I understand it, anyways.
Brownian motion does not need a gradient.
If you do set up a system with a concentration gradient, particles will tend to flow from the side with higher concentration to the side with lower concentration just because there is more of them on the higher side to randomly walk to the lower side. For example, set up a system with lots of particle A on one side and little of it on the other. Here [A] is “concentration of A”, P is “probability of”.
high [A] | low [A]
P finding any given A: 0.9 | 0.1
P A staying on the same side: 0.5 | 0.5 } random
P A going to the other side: 0.5 | 0.5 } walk
After some time:
P finding any given A 0.9*0.5+0.1*0.5 | 0.1*0.5+0.9*0.5
or 0.5 | 0.5
If you aren’t actively doing something to maintain the gradient, Brownian motion (diffusion) will destroy that gradient and the concentration will be the same on both sides.
Edits: Formating
Because if there are more things on the left than on the right, the chance of a left item going right is bigger than the chance of a right item going left. Not per item, but accumulated over everything.
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