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Topic: Confused by easy equilibrium!  (Read 3611 times)

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Offline smghz

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Confused by easy equilibrium!
« on: December 01, 2016, 04:49:40 PM »
Hi!

I was wrestling a bit with these questions while studying equilibrium.

  • Why is concentration so important?
  • Why isn't equilibrium dependent on pure solids or liquids? Say it is the thermal reaction of calcium carbonate. Is not the solid used up? Isn't at least part if not all of the solid utilized for the reaction? And if so, why isn't that affecting the equilibrium?
  • And what is really the difference between the amount of something reacting and the concentration, and which is more important?

thanks.

Offline AWK

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Re: Confused by easy equilibrium!
« Reply #1 on: December 01, 2016, 07:27:58 PM »
AWK

Offline smghz

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Re: Confused by easy equilibrium!
« Reply #2 on: December 01, 2016, 08:49:22 PM »
1. Activity is important!
https://www.wikiwand.com/en/Chemical_equilibrium

So my question is, say we go back to that calcium carbonate example. If we added, like, 15 moles of CaCO3, why does the equilibrium position still remain the same, unlike say we add gaseous CO2??? thanks.

Online Borek

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Re: Confused by easy equilibrium!
« Reply #3 on: December 02, 2016, 03:02:34 AM »
Imagine reaction taking place in the solution. Added gaseous CO2 dissolves and its concentration changes. Adding solid to the mixture doesn't change its concentration in the liquid phase.
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Offline smghz

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Re: Confused by easy equilibrium!
« Reply #4 on: December 02, 2016, 09:29:43 AM »
Imagine reaction taking place in the solution. Added gaseous CO2 dissolves and its concentration changes. Adding solid to the mixture doesn't change its concentration in the liquid phase.

But my question cycles back to the consumption of the solid. Isn't some of the solid being used? Why isn't that affecting the concentration, esp. that some of the solid molecules now become gas molecules??? Also, doesn't adding water lead the water itself to become part of the aqueous solution?

Online Borek

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Re: Confused by easy equilibrium!
« Reply #5 on: December 02, 2016, 01:54:46 PM »
And how do you define concentration of the solid, so that you know it changes? Concentrations is a property of the homogeneous mixture, mixture of solid and liquid is not homogeneous.

Reaction takes place only on the solid surface, it is enough that the surface exist for the solid to be part of the equilibrium. Sure, the higher the surface the faster the reaction is, but for equilibrium both forward and reverse reactions are important - and both speeds are dependent on the surface in exactly the same way, so it cancels out.
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Offline smghz

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Re: Confused by easy equilibrium!
« Reply #6 on: December 02, 2016, 06:53:24 PM »
And how do you define concentration of the solid, so that you know it changes? Concentrations is a property of the homogeneous mixture, mixture of solid and liquid is not homogeneous.

Reaction takes place only on the solid surface, it is enough that the surface exist for the solid to be part of the equilibrium. Sure, the higher the surface the faster the reaction is, but for equilibrium both forward and reverse reactions are important - and both speeds are dependent on the surface in exactly the same way, so it cancels out.

That makes a heck of a LOT more sense! So when we have a pure solid or liquid, no matter the amount, the equilibrium position will remain the same because they don't change in concentrations (and in corollary the rate of collisions between molecules). Even if we increase the amounts, still there would be an equilibrium at the same position because they aren't affecting the concentrations, unlike for gases and aqueous solutions whose concentrations can change and therefore shift the equilibrium position to the right or left depending on Le Chatelier's principle.

THANKS SO MUCH!!!

Offline Scitalk

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Re: Confused by easy equilibrium!
« Reply #7 on: December 10, 2016, 05:41:29 PM »
I'll tackle the first question. Concentration is IMPORTANT. The concentration of a reactant directly effects the rate and shift of the reaction at equilibrium. Chemical reactions have a equilibrium constant K. That value is CONSTANT. The products divided by the reactants equals this constant K. If a concentration value increases then other concentration values must decrease to maintain that k constant. This pretty much sums it up! Haha

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