[MaxShlochz]

Hello, this is my first post here and I'm pretty sure that this question has been answered before
but I couldn't really find it in a proper manner on the web.

How come the only thing that can change the equilibrium constant is the temperature and not
the additions of a concentration/concentrations or the pressure?

I know that the equilibrium constant is ratio between the rate of the reaction going forward divided by the rate backwards BUT if we increase pressure or add more concentration - Won't we be getting more bumps and because of that, the rate of the reaction forwards/backwards will increase?

This has been bugging me for quite a while, I hope I'll get the sufficient help I need.
Any help will be much obliged.

[magician4]

How come the only thing that can change the equilibrium constant is the temperature and not
the additions of a concentration/concentrations or the pressure?

well, you'll have to take that with a pinch of salt: "within reasonable limits" (for example, you might wish to become not too diluted - a handfull of molecules make for bad statistics - or too concentrated either)
but for most practical purposes: yes, that's the situation

I know that the equilibrium constant is ratio between the rate of the reaction going forward divided by the rate backwards (...)

though the LMA often is introduced this way: no, that's not completely correct
instead, correct definition would be that the chemical potentials of forward- and backwards reaction in equilibrium are identical, and their difference hence becomes zero.

BUT if we increase pressure or add more concentration - Won't we be getting more bumps and because of that, the rate of the reaction forwards/backwards will increase?

yes, we would . BUT ...
.. at equilibrium (! and only then, and only for this situation the K's are meaningfull) this would mean, that the number of product forming processes would be , let's say for example, doubled - but so would the reverse processes, as at equilibrium there also would be the same proportionally increased amount of reverseable product particles present.
so, instead of (let's say , just to clarify what I'm talking about) a hundred million processes per millisecond going both ways at low conc. (that's what "equilibrium" means) , we suddenly had like two hundred million of those processes - but, therefore, still no net gain in either direction, and hence still the same K-value


regards

Ingo

[MaxShlochz]

yes, we would . BUT ...
.. at equilibrium (! and only then, and only for this situation the K's are meaningfull) this would mean, that the number of product forming processes would be , let's say for example, doubled - but so would the reverse processes, as at equilibrium there also would be the same proportionally increased amount of reverseable product particles present.
so, instead of (let's say , just to clarify what I'm talking about) a hundred million processes per millisecond going both ways at low conc. (that's what "equilibrium" means) , we suddenly had like two hundred million of those processes - but, therefore, still no net gain in either direction, and hence still the same K-value

So what you're saying is, that because a concentration of some molecule/s has doubled, it will start bumping into each other and creating (Let's say we've added to the concentration of the reactants)
more product but at the same time the products will create more reactants and because of that the equilibrium constant does not change at all.

Did I more or less understood you correctly?

One more clarification about this if I may I ask, I'm guessing that the same happens when we add pressure? (only with the addition of pressure, it will go to the place which has less moles of molecules)

[magician4]

Did I more or less understood you correctly?

yes, you did

One more clarification (...)

talking about the influence of pressure, the situation indeed is next to identical. Only difference would be, that - as "pressure" mostly is relevant in gas phase reactions - the general behaviour of gases ref. pressure variations has to be taken into account , too.
this means, that you'd have to include the gas laws into the picture  ( mostly this is reduced to pV = n*R*T), resulting in a gas - related equilibrium constant K_p

However, even with gases the law of mass action might be expressed concentration related, and there is a respective K_c for those situations, too.

those two constants K_p and K_c are interconnected by the equation

[tex] K_p = K_c (RT)^{\Delta n} [/tex]

(for more detailled information pls. take a look here: link )


regards

Ingo

[MaxShlochz]

Well, you definitely gave me something to think about, especially about that formula, it's the first time I'm seeing it so thanks for the help and the link you gave at the end, you've been most helpful.