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Offline mikhailconrad

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Chemical Equilibria
« on: February 11, 2016, 10:06:23 AM »
I have a small question about Chemical Equilibria, why exactly do we multiply the concentration/activities of the Reactants/Products? I remember watching a Khan Academy video saying something that it's really complex and that, if I remember correctly, it has something to do with the probability of finding these ions together; I assume the same way you only have a 25% (50%*50%) chance to get two heads in a row with coin tosses.

Is this intuition anywhere near close? Am I ever going to get a better explanation of this with higher maths and chemistry?

Thanks in advance for any answers.

Offline mikasaur

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Re: Chemical Equilibria
« Reply #1 on: February 11, 2016, 11:44:53 AM »
Conceptually you can really, really simplify why an equilibrium quotient [itex]Q[/itex] is what it is. Let's say we have the reaction:

aA + b::equil:: cC + dD

In order for this to proceed in the forward reaction some A and B have to meet. The more A and B there is the more likely their chances of meeting. It is like your coin analogy. A better (but much weirder) analogy might be two rooms with objects on the floors.

Let's say in one room you have a bunch of big dinner plates on the floor and in the other room you have a bunch of tea plates. These can represent our concentrations of A and B. The more plates, the higher the concentration!

Imagine tossing a coin to some random spot on the floor. The more plates the higher the chance of hitting a plate. Let's say that a "reaction" occurs when you walk to one room, randomly toss a coin and hit a plate, then walk to the other room and do the same thing. If you miss in either room a reaction doesn't occur.

If one room is half covered in plates and the other only one third, your chances for a "reaction" occurring are [itex]\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}[/itex]. There's the multiplication! Change the "concentration" of plates and you change the chances of a "reaction" occurring.

Make sense?

Now if you want to go into the thermodynamic derivation of reaction quotients we can do that. It involves Gibbs energy and logarithms and math and is a ton of fun! :P
Or you could, you know, Google it.

Offline Borek

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Re: Chemical Equilibria
« Reply #2 on: February 11, 2016, 11:49:42 AM »
This is more or less equivalent to the kinetic approach to the equilibrium, where the speed of the reaction is proportional to concentrations in correct powers of the reacting compounds (molecules have to hit each other, and the probability of the collision can be estimated from the models mikasaur refers to), and to reach equilibrium we need to get to the point where the forward reaction speed equals the backward reaction speed. Sadly it works nicely only for very simple reactions, once you know a bit kinetics it is obvious this approach must be off in general. Then it is all thermodynamics and actually fact that the kinetic approach works is quite coincidental.
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Offline mikhailconrad

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Re: Chemical Equilibria
« Reply #3 on: February 12, 2016, 02:25:44 AM »
Now if you want to go into the thermodynamic derivation of reaction quotients we can do that. It involves Gibbs energy and logarithms and math and is a ton of fun! :P

Thanks! I'll bet it is, but I'll have to pass for now. We haven't gotten to thermodynamic equations yet, so I have no idea what Gibbs energy even is.

Then it is all thermodynamics and actually fact that the kinetic approach works is quite coincidental.
That's hilarious.

Offline aga

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Re: Chemical Equilibria
« Reply #4 on: February 16, 2016, 03:31:26 PM »
Kinetics is basically the chances of two reactable bits finding each other.

In the simplest terms :-

The chance of a young guy finding a girl on a desert island are basically none (=0 kinetic probability).
No matter how 'active' he would like to be (=thermodynamics) it will simply not happen.

Put the same guy in a higher concentration of women (e.g. a girl's high school end-of-term party) and he has a much higher chance encountering a 'favourably reactive' female.

Molecules/atoms face exactly the same problem, which is solved in pretty much the same way.
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