[T]

Hello,

I have been learning about equilibrium constant and based on what I have read the constant displays the ratio/relationship between the concentration of products and reactants. Why does the constant stay the same even when the pressure or concentration changes? If the concentration of A increases in A+B ::equil::C, won't the ratio change? Also what do you use the equilibrium constant for? Do you find the concentration of C when you have A and B?

Thanks

[Borek]

I have been learning about equilibrium constant and based on what I have read the constant displays the ratio/relationship between the concentration of products and reactants

Please don't call it just a 'ratio', as it is misguiding. It is better to speak about the reaction quotient, Q. For the reaction

aA + bB cC + dD

reaction quotient is

[tex]Q = \frac {[A]^a[ B]^b}{[C]^c[D]^d}[/tex]

When the reaction is at equilibrium Q=K.

Why does the constant stay the same even when the pressure or concentration changes?

Simple answer - because that's the way it works. You can explain it using thermodynamics.

If the concentration of A increases in A+B C, won't the ratio change?

Initially it will, but then the reaction will shift till it gets to a new equilibrium, where Q=K. There is nothing wrong with Q≠K, it is just not a stable situation.

Also what do you use the equilibrium constant for? Do you find the concentration of C when you have A and B?

Exactly, I guess that's what you will be learning about in the next weeks.

[Big-Daddy]

Why does the constant stay the same even when the pressure or concentration changes?

As Borek said it "turns out" to be the case but you can demonstrate it without too much difficulty: http://en.wikipedia.org/wiki/Chemical_equilibrium (of course, some of the concepts on the way to proving it, are difficult, but the maths is straightforward, if you will accept these bare bones of a rigorous proof)

Also what do you use the equilibrium constant for? Do you find the concentration of C when you have A and B?

Not just that. You can predict the equilibrium constants of hypothetical reactions with the equilibrium constants of known reactions. And you could find the equilibrium concentrations of C (and A and B) from the initial concentrations of A and B (taking your example), using the value of the equilibrium constant and some reaction stoichiometry.

[T]

Thanks Borek and Big-Daddy,

A question for Big-Daddy:
Big-Daddy

Not just that. You can predict the equilibrium constants of hypothetical reactions with the equilibrium constants of known reactions. And you could find the equilibrium concentrations of C (and A and B) from the initial concentrations of A and B (taking your example), using the value of the equilibrium constant and some reaction stoichiometry.

Could you expand more on the part where you can use initial concentration to find the equilibrium concentration of C?

Also could someone explain to me why solids and liquids are not included when calculating the constant?

And could someone tell me why you would need Kp instead of Kc? If it just depends on whether you're given pressure or concentration then write that there are no advantages of using either.

Thanks

[Big-Daddy]

Could you expand more on the part where you can use initial concentration to find the equilibrium concentration of C?

If you have trouble with any of the steps that follow, think about it and if you don't get it then pinpoint the step and I will try to explain.

First let's take a single-reaction case, aA+bB  cC+dD for generality. Let us define n_react(A) as the amount of A which reacts and n_react(B) as the amount of B which reacts; n₀(P) as the initial number of moles of species P, where P=A,B,C,D etc.; n_eq(P) as the equilibrium number of moles of species P. Our task is to express the equilibrium constant K, not in terms of the equilibrium numbers of moles of A,B,C,D but rather in terms of the equilibrium number of moles of any one of them, along with the initial numbers of moles of all of them.

Self-evidently n_react(A) = n₀(A) - n_eq(A) and analogous for B (the reactants).
Now, we can also say from the reaction stoichiometry that n_react(B) = b/a * n_react(A).  Combining with the above, (b/a)*(n₀(A)-n_eq(A))=n₀(B)-n_eq(B). Let's now decide we will work first calculate the equilibrium number of moles of A first. Then, we should rearrange the expression we just got for n_eq(B) in terms of n₀(B), n₀(A) and n_eq(A).

Next, C and D. n_eq(C) = n₀(C) + n_react(A)*(c/a) from stoichiometry. Analogous equation for D. Substitute in n_react(A) = n₀(A) - n_eq(A), and you get e.g. n_eq(C) = n₀(C) + (n₀(A)-n_eq(A))*(c/a). Do the same for n_eq(D).

Now we move to work with K instead.

K = a(C)^c*a(D)^d / (a(A)^a*a(B)^b) where a(P) is activity of P
a(P) ≈ c(P)/c° for species P, if it is a solute, or p(P)/p° if it is a gas (under certain circumstances known as ideality)
So K = (c(C)^c*c(D)^d) / (c(A)^a*c(B)^b) * (c°)^(a+b-c-d) (or analogous for pressures)

And now you convert from your numbers of moles (obtained above) to concentrations or partial pressures and substitute in, leaving you with a final expression for K in terms of n_eq(A) as well as n₀(A), n₀(B), n₀(C), n₀(D) and the stoichiometric coefficients. You solve this expression (numerically if need be) to find n_eq(A). Then you use the same relationships you found and substituted into K to then find n_eq(B), n_eq(C), n_eq(D). And then you're done! You can convert these to concentrations or partial pressures after that if you need.

Also could someone explain to me why solids and liquids are not included when calculating the constant?

The equilibrium constant is really in terms of activities, and activities of solids and liquids are taken to be unity. Thermodynamics can be used to show why this should be approximately the case in some situations.

And could someone tell me why you would need Kp instead of Kc? If it just depends on whether you're given pressure or concentration then write that there are no advantages of using either.

There is no fundamental difference. a(P) ≈ c(P)/c° if you are dealing with solutes or a mixture of liquids and a(P) ≈ p(P)/p° if you are dealing with gases. This leads to K=K_p if the reaction only involves gases and K=K_c if the reaction only involves solutes or a mixture of liquids. Only this K, the true thermodynamic equilibrium constant, is actually necessarily an equilibrium constant.

If a reaction involved both solutes and gases, you could not merely take one or the other and establish an equilibrium constant with them (it would not be a constant).

[Borek]

If you have trouble with any of the steps that follow

You have a tendency to absurdly complicate things. No need for overgeneralization that just makes things confusing. Why not just calculate pH of the 0.1 M acetic acid solution?

CH₃COOH CH3COO^- + H⁺

Equilibrium constant (which happens to be called dissociation constant) is

[tex]K_a = \frac {[CH_3COO^-][H^+]}{[CH_3COOH]} = 1.8 \times 10^{-5}[/tex]

Initially there was 0.1 moles per liter of acetic acid. Some of it dissociated, producing CH3COO^- and H⁺. The equilibrium concentration of undissociated acetic acid is lower than its initial (analytical) concentration. Initially we had 0.1 M, but some of it dissociated. How much? Exactly as much as there were CH3COO^- produced, so from stoichiometry

[CH₃COOH] = 0.1 M - [CH₃COO^-]

Also from the stoichiometry we know amounts of [CH3COO^-] and [H⁺] produced during dissociation are identical, so [CH3COO^-]=[H⁺].

So we have

[tex]\frac {[CH_3COO^-][H^+]}{[CH_3COOH]} = \frac {[CH_3COO^-][H^+]}{0.1 - [CH_3COO^-]} = \frac {[H^+][H^+]}{0.1 - [H^+]} = 1.8 \times 10^{-5}[/tex]

Solve for [H⁺] - this is just a simple math now. There will be two solutions, but one is negative, so it can't be right. The other is the correct one.