Exploring Equilibrium Reactions: Questions & Answers

In summary, water concentration remains constant, but the concentration of solid substances and precipitates does not. We calculate the concentration of a substance by multiplying its concentration by the concentration of the substance's reactants and products. The importance of calculating the concentration of Kw is unknown.
  • #1
Misr
385
0
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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  • #2
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.
 
  • #3
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.
 
  • #5
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

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

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

[tex]K_w = a_{H^+}a_{OH^-}[/tex]

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).

--
 

Related to Exploring Equilibrium Reactions: Questions & Answers

1. What is an equilibrium reaction?

An equilibrium reaction is a chemical reaction in which the concentration of reactants and products remain constant over time, meaning the rate of the forward reaction is equal to the rate of the reverse reaction.

2. How do you determine the direction of an equilibrium reaction?

The direction of an equilibrium reaction is determined by the relative concentrations of the reactants and products. If the reactants are more concentrated, the reaction will favor the formation of products, and vice versa.

3. What factors can affect the equilibrium of a reaction?

The equilibrium of a reaction can be affected by changes in temperature, pressure, and concentration of reactants and products. Adding or removing a reactant or product, or changing the volume or pressure of the system, can also shift the equilibrium.

4. How do you calculate the equilibrium constant of a reaction?

The equilibrium constant, K, can be calculated by dividing the product of the concentrations of the products by the product of the concentrations of the reactants, each raised to the power of their respective coefficients in the balanced chemical equation.

5. Can the equilibrium constant of a reaction be changed?

No, the equilibrium constant of a reaction is a constant value at a given temperature. However, the equilibrium position can be shifted by changing the conditions of the reaction, such as temperature or concentration of reactants and products.

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