Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: xshadow on October 15, 2015, 05:51:49 PM
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Hi!!
First of all sorry for my bad english but it isn't my 1st language!
I need some help to understand the different meaning of [itex]K(T)[/itex] and [itex]Kc[/itex] where:
K(T)= K_c* [y( C)* y(D)]/[y(A)*y(B)] = K_c * k_y
where:
[itex] y(X)[/itex] is the activity coefficient of the X-th species
[itex]K_c [/itex] is the usual equilibrium constant with concentrations.
So looking this formula i don't understand if K(T) is a CONSTANT for a some reaction at a T=T_0 temperature or if his value change ?? same doubt for K_c !!
For example, for the reaction at the temperature [itex] T_0 = 298K [/itex] :
CH3COOH+H2O ::equil:: CH3COO- + H30+
1)the K(T) (the thermodynamics constant) can change his value?? for example due to the ionic strenght and so due to the different values of the activity coefficients??
2) and K_c can change his value for the same reason?? (different ionic strenght ---> different activities coefficients)
At last what is the main differences (of meaning) between these two equilibrium constant??
If someone can help me...
Thanks :)
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I'm not entirely familiar with the terminology here but let's have a try.
First, let's introduce the concept of the reaction quotient Q. For a reaction
A + B ::equil:: C + D
Q = [C][D]/[A][B ] (This may be at any point in the reaction, not necessarily equilibrium)
The equilibrium constant Kc is equal to the value of Q at equilibrium, Qeq.
This is strictly true only at high dilution, when activity equals concentration. Let us call this equilibrium constant Kc∞.
Now we could define Kc ≡ Kc∞, and this is a constant. Then we would say that at higher concentrations, when activity differs from concentration, Qeq may be different from Kc. Kc is constant, but Qeq is variable.
Alternatively (and I don't know which practice is currently fashionable) we could define Kc ≡ Qeq under all conditions, and then say that Kc, which is variable, differs from Kc∞, which is constant.
Now let's consider it in terms of activities, α. We define
Qα = αCαD/αAαB
and Kα ≡ Qαeq = QeqγCγD/γAγB
This is always true, from the definition of activity - it is that quantity which behaves as concentration ideally should. Kα is constant.
At high dilution γ = 1, so Kα = Kc∞
If you take the second alternative above, where Kc = Qeq and is variable, then
Kα = KcγCγD/γAγB
and Kα corresponds to your K(T).
I hope this helps a little. The key point is that the equilibrium constant in terms of activities is a true constant - that's what activity means.
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Thanks!! :)
I think to have understood the point:
The general expression for a reaction 1A+1B ::equil:: 1C+1D is:
K(T)= {a(C)*a(D)}/{a(A)*a(B)}= {[C]*[D]}/{[A]*[ B]}* Ky= Kc*ky
Here K(T) is a costant for a reaction at a certain temperature: in other words the expression {a(C)*a(D)}/{a(A)*a(B)} has a constant value for a reaction.
What is change is Kc and Ky (the activity coefficients )...
When Ky changes, i.e. the activity coefficients change(due to ionic strenght,for example), the term Kc has to change in order to guarantee the numeric value espressed by K(T)= {a(C)*a(D)}/{a(A)*a(B)} ,which is a constant and must be always respected.
When Ky=1 :rarrow: K(T)=Kc and in this particular case Kc has the usual value that we find generally in the books for that reaction...for example for CH3COOH at 25*C has the common value of 1,75*10^(-5)