Why Are Water and Solids Excluded from Equilibrium Constant Calculations?

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SUMMARY

The discussion clarifies why water and solids are excluded from equilibrium constant calculations, specifically in the context of the equilibrium constant Kc. Water's concentration remains effectively constant at approximately 55.5 moles/liter, making its inclusion unnecessary for most calculations. Solids, on the other hand, do not have a meaningful concentration in reactions as they react only at their surfaces. The conversation also touches on the distinction between concentration and activity, particularly in gaseous states, and the importance of the water ion product (Kw) in relation to these concepts.

PREREQUISITES
  • Understanding of equilibrium constants (Kc) in chemical reactions
  • Knowledge of stoichiometry and concentration calculations
  • Familiarity with the concept of activity in chemistry
  • Basic principles of acid-base chemistry, including the water ion product (Kw)
NEXT STEPS
  • Research the concept of activity coefficients in chemical equilibria
  • Study the implications of the water ion product (Kw) in acid-base chemistry
  • Explore the differences between concentration and activity in chemical reactions
  • Learn about Ostwald's law and its application in dissociation calculations
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Chemistry students, educators, and professionals involved in chemical equilibrium, acid-base reactions, and those seeking to deepen their understanding of reaction kinetics and thermodynamics.

Misr
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1-The concentration of water,solid substances and precipates should not appear in the equilibrium constant equation Kc , why?I don't understand.Please don't tell me that because their concentration remains constant whatever their quantities. because I have read these words tens of times and I don't really understand them.

2-according to my textbook " if the reactants and the products are in the gasoeus state,the concentration is expressed by their partial pressure" What does this mean?

3-CH3COOH <----->CH3COO- + H3O+

let a be the no. of dissociated moles.So if we want to calculate the concentration of CH3COOH why should we say its remaining concentration=C-a,where C is the orginal concentration of CH3COOH before dissociation?.

4-H2O <---->H+ + OH-
Kc= [H+] [OH-]/[H2O]=10^-14
Kw=[H+] [OH-] = 10^-14
how could the two equations have the same value?why we are totally neglecting water?and what is the importance of calculating Kw?

I've asked a question here here but it made no sense so please don't repeat the same answer.
http://answers.yahoo.com/question/i...sH4G60Dty6IX;_ylv=3?qid=20101210130754AASJDiN
Help me please I'm very confused!
Thanks in advance.
 
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Misr said:
1-The concentration of water,solid substances and precipates should not appear in the equilibrium constant equation Kc , why?I don't understand.Please don't tell me that because their concentration remains constant whatever their quantities. because I have read these words tens of times and I don't really understand them.

Concentration of water is almost constant - it is about 55.5 moles/liter. Even if water reacts with some of the substances present, its concentration rarely changes by more than tenths of percent. We rarely know equilibrium constants with accuracy high enough for this change to make a difference. So in most cases we can safely assume water concentration to be constant - but it is not always true.

Solids are not dissolved, they are reacting only on the surface, it is enough that they are present to take part in the reaction. Concentration of the solid - which is separate from water - doesn't make much sense.

2-according to my textbook " if the reactants and the products are in the gasoeus state,the concentration is expressed by their partial pressure" What does this mean?

Not concentration, but activity, in fact each time we talk about reaction quotient we use activities, not concentrations. As a first approximation concentration and activity have the same value.

Do you know what partial pressure is?

3-CH3COOH <----->CH3COO- + H3O+

let a be the no. of dissociated moles.So if we want to calculate the concentration of CH3COOH why should we say its remaining concentration=C-a,where C is the orginal concentration of CH3COOH before dissociation?.

Stoichiometry. Imagine you started with 1 mole of acetic acid and 0.1 moles dissociated - obviously 1-0.1=0.9 moles are left. Now imagine it happened in a known volume of solution - you start with C=1/V, concentration of dissociated acid is a=0.1/V, concentration of the left acid is (1-0.1)/V=C-a.

4-H2O <---->H+ + OH-
Kc= [H+] [OH-]/[H2O]=10^-14
Kw=[H+] [OH-] = 10^-14
how could the two equations have the same value?why we are totally neglecting water?and what is the importance of calculating Kw?

They don't have the same value. See water ion product page.
 
Actually this is very helpful
1-
Concentration of water is almost constant
Yes.but if what is the problem if it is constant? why don't we substitute with the concentration of water as 55.5 moles/liter ?and what is the relation between calculation of Kw and assuming water concentration constant?

2-
Not concentration, but activity, in fact each time we talk about reaction quotient we use activities, not concentrations. As a first approximation concentration and activity have the same value.

Do you know what partial pressure is?
yes i know.Do u mean by "activity" the rate of chemical reaction ?
As a first approximation concentration and activity have the same value.
how?can u give me an example?

3-
Stoichiometry. Imagine you started with 1 mole of acetic acid and 0.1 moles dissociated - obviously 1-0.1=0.9 moles are left. Now imagine it happened in a known volume of solution - you start with C=1/V, concentration of dissociated acid is a=0.1/V, concentration of the left acid is (1-0.1)/V=C-a.
got it now!

4-
They don't have the same value. See water ion product page.
as usual..a mistake in my book.

Thanks very much.
 
Misr said:
Yes.but if what is the problem if it is constant? why don't we substitute with the concentration of water as 55.5 moles/liter ?and what is the relation between calculation of Kw and assuming water concentration constant?

Not sure what you are asking about. We ignore water concentration to make calculations easier. It doesn't mean it is not there - just like in the case of water ion product, we move it to the equilibrium constant. This way we have one number less to worry about, it makes calculations easier.

yes i know.Do u mean by "activity" the rate of chemical reaction ?

No, rate is rate, activity is activity. For example

a_{H^+} = f_{H^+}[H^+]

activity of H+ equals its concentration times activity coefficient. For water ion product

K_w = a_{H^+}a_{OH^-}

For diluted solutions activity coefficients equal 1, for not too concentrated solutions (say below 0.1M) they are identical for ions of the same charge (so fH+=fOH-) and are less than 1. See ionic strength and activity coefficients. Please note that this is just an approximation of what is happening in reality - better one than using just concentrations, but failing in more concentrated solutions. In fact despite over 100 years of efforts we still don't have a reasonable theory allowing calculation of activities of ions in more concentrated solutions.

Misr said:
another question:
In Ostwald law, sometimes we assume (a) as no. of dissociated moles, sometimes we assume (a) as degree of dissociation which equals no. of dissociated moles/the total no. of moles of a substance before dissociation
http://www.pinkmonkey.com/studyguides/subjects/chem/chap12/c1212701.asp
so, what is (a) ?

There is some problem with formatting on the page, seems to me like whenever they use "a" they in fact mean α (degree of dissociation).

--
 

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