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### Topic: Thermodynamic equilibrium constant K(T) and Kc  (Read 1717 times)

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#### xshadow

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• Mole Snacks: +1/-0 ##### Thermodynamic equilibrium constant K(T) and Kc
« on: October 15, 2015, 05:51:49 PM »
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 $K(T)$ and $Kc$ where:

K(T)= K_c* [y( C)* y(D)]/[y(A)*y(B)] = K_c * k_y
where:
$y(X)$ is the activity coefficient of the X-th species
$K_c$ 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 $T_0 = 298K$ :
CH3COOH+H2O 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 « Last Edit: October 16, 2015, 03:20:37 AM by Borek »

#### mjc123

• Chemist
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• Mole Snacks: +282/-12 ##### Re: Thermodynamic equilibrium constant K(T) and Kc
« Reply #1 on: October 16, 2015, 05:23:51 AM »
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 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 = αCαDAαB
and Kα ≡ Qαeq = QeqγCγDAγB
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 = KcγCγDAγB
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.

#### xshadow

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• Mole Snacks: +1/-0 ##### Re: Thermodynamic equilibrium constant K(T) and Kc
« Reply #2 on: October 16, 2015, 12:48:23 PM »
Thanks!! I think to have understood  the point:

The general expression  for a reaction 1A+1B 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 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)

« Last Edit: October 16, 2015, 02:18:41 PM by Borek »