Chemical Equilibrium Question

  • #1

Main Question or Discussion Point

Hey guys,

I have a question regarding the process by which a reaction returns to equilibrium.

If a reaction's reaction quotient is greater than its equilibrium constant, it will shift towards the reactants side of the equation to return to equilibrium. How exactly does this happen? Does the forward reaction stop, or does the rate of the reverse reaction increase so reactants are being produced quicker than products? Is there ever a point where products or reactants are not being produced at all?
 

Answers and Replies

  • #2
397
21
Equilibrium does not mean things stop moving or reacting or what not. You have to look at chemical processes as particles which are constantly vibrating, moving, deforming, colliding etc. It is pretty much chaos. When you see an illustration or video of two molecules neatly coming together and forming bonds or what-have-you, you must understand that it is only a model used to describe an interaction. In reality, those two molecules may have collided a zillion times without the proper orientation or energy to form bonds and on the zillionth and one collision they have enough energy and the proper orientation to actually form energetically favorable bonds. Taking this view point makes it pretty intuitive that rates of reactions rely heavily on concentrations of reactants.

Now as far as equilibrium goes, it means that the rate of the reverse reaction exactly equals the rate of the forward reaction. There are still interactions happening, with bonds breaking and forming, its just that if you measure concentrations of products and reactants they will have not net change. Weak acids for instance don't just dissociate to a specific amount and just sit around in a predictable ratio. Its more like the acidic protons are zooming throughout the whole system constantly associating and dissociating with anything that can accommodate them. Its just that the net result is one where there is no more change in concentration of acid and conjugate base. But if you follow on molecule around, hypothetically, it is constantly being protonated and deprotonated.
 
  • #3
Borek
Mentor
28,398
2,800
To add to Yanick's answer - this is so called kinematic approach to equilibrium, and equilibrium is dynamic - there are both forward and forward reaction taking place, but their speeds are equal.
 

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