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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: smghz on December 01, 2016, 04:49:40 PM

Title: Confused by easy equilibrium!
Post by: smghz on December 01, 2016, 04:49:40 PM
Hi!

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


thanks.
Title: Re: Confused by easy equilibrium!
Post by: AWK on December 01, 2016, 07:27:58 PM
1. Activity is important!
https://www.wikiwand.com/en/Chemical_equilibrium
Title: Re: Confused by easy equilibrium!
Post by: smghz 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.
Title: Re: Confused by easy equilibrium!
Post by: Borek 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.
Title: Re: Confused by easy equilibrium!
Post by: smghz 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?
Title: Re: Confused by easy equilibrium!
Post by: Borek 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.
Title: Re: Confused by easy equilibrium!
Post by: smghz 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!!!
Title: Re: Confused by easy equilibrium!
Post by: Scitalk 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