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Topic: Equilibrium constant, Kc, question  (Read 1196 times)

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Offline hipas

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Equilibrium constant, Kc, question
« on: May 23, 2020, 09:53:13 AM »
Hi!

A friend of mine insists that there are cases where a reactant or product in a liquid state is used in the equilibrium constant calculation ( the equilibrium constant Kc, Kc = [products]/[reactants]). And I believe that no, a liquid or a solid do not take part in the calculation of Kc, ever.

I've read in many places, that a pure liquid or a pure solid do not take part in the calculation of Kc. So that refers to any reactant or product that has an (l) or an (s) after their chemical formula in the chemical equation, right? We can't have a represention of "chemical formula (l)" without the substance being pure, can we? For example if the substance is potassium chloride, KCl, if KCl(l) appears in a chemical equation, the potassium chloride is pure because if it wasn't the chemical formula wouldn't be KCl, it would be a solution, not a substance. So any product or reactant with an (l) after is a pure liquid and will not take part in the calculation of Kc. Am I correct or nah?

Thank you!

Offline Borek

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Re: Equilibrium constant, Kc, question
« Reply #1 on: May 23, 2020, 10:52:22 AM »
alcohol(l) + acid(l) ::equil:: ester(l) + water (l)
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Offline hipas

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Re: Equilibrium constant, Kc, question
« Reply #2 on: May 23, 2020, 11:17:51 AM »
alcohol(l) + acid(l) ::equil:: ester(l) + water (l)

Hi! I'm trying to find an exercise where where they determine the Kc of this type of reaction (esterification reaction?) because I wanted to know how different is the concentration of a acid(l) from a acid(aq). And how that H2O(l) is done. How any (l) reactants and products concentrations are done, I guess. Thank you

Offline Borek

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Re: Equilibrium constant, Kc, question
« Reply #3 on: May 23, 2020, 01:12:55 PM »
how different is the concentration of a acid(l) from a acid(aq)

How is the concentration defined?

If you take 1 mole of acetic acid and fill it up to 1 L with water, what will be the acid concentration?

If you take 1 mole of acetic acid and fill it up to 1 L with ethanol, what will be the acid concentration?

Does it qualify as a "pure substance" in each case?
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Offline hipas

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Re: Equilibrium constant, Kc, question
« Reply #4 on: June 10, 2020, 01:02:43 PM »
Thank you very much for replying :) :) :) :)

Molar concentration is moles of solute divided by volume of solution, so in solution 1 we'd have 1 mol/L of an aqueous acetic acid solution because the solvent would be water and solution 2 we'd have 1 mol/L of an alcoholic acetic acid solution, because the solvent is alcohol?

Is it because between the alcohol, acid, ester and water in that reaction there isn't one component that isn't in such a bigger proportion (like water in an acid-base reaction) and so all of them are going to take part in the calculation of the equilibrium constant? And so that requires that the alcohol and acid are very concentrated, occupy more than 50% of volume in a solution and so are actually solvents themselves?
« Last Edit: June 10, 2020, 01:42:32 PM by hipas »

Offline Borek

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Re: Equilibrium constant, Kc, question
« Reply #5 on: June 10, 2020, 03:51:45 PM »
There is no strict definition as to when something becomes a solvent, this is often a matter of convention. Commercial nitric acid is around 68% w/w, so technically we should rather speak about solution of water in the acid, yet nobody does that.
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