Kc is the constant of equilibrum

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In summary, Kc is the constant of equilibrum for a certain chemical reaction. For example, if the chemical reaction has the following formula: A+2B->C+3D, then Kc=([C].[D]3)/([A].[B]2). This is what they are teaching us at least, but i find it not really logical, for may reasons, here is one. Suppose we multiply the whole formula by 2: 2A+4B-->2C+6D. The value of Kc will get squared, but this seems wrong since both fomulas are for the same reaction! Anyone can explain what is happening please? Thanks in advance.
  • #1
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Kc is the constant of equilibrum for a certain chemical reaction.
For example, if the chemical reaction has the following formula :
A+2B->C+3D
Then Kc=([ C ].[ D ]3)/([ A ].[ B ]2)
(Where [ A ] means the molarity of the compound A)
This is what they are teaching us at least, but i find it not really logical, for may reasons, here is one.
suppose we multiply the whole formula by 2
2A+4B-->2C+6D
(the equation was edited after the notice of Mike)
The value of Kc will get squared, but this seems wrong since both fomulas are for the same reaction !

Anyone can explain what is happening please ?
Thanks in advance :smile:
 
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  • #2
That should be a 6D and not a 3D in your second reaction.

I suspect that to be your problem.
 
  • #3
Have you been given the relation between free energy and equilibrium in your class? If not, it's all going to look like an arbitrary mystery; if you have, recall that K is specific for the reaction as written, and recognize that the free energy for the reaction as you have rewritten it for twice the reactants and products is also doubled.
 
  • #4
Have you been given the relation between free energy and equilibrium in your class
Nop ! Can you please explain more ?
free energy for the reaction as you have rewritten it for twice the reactants and products is also doubled
As i see the value of Kc will be squared not doubled, is that what you mean ?
 
  • #5
Welcome to the wonderful world of chemical thermodynamics --- jumping in in the middle and working backward probably isn't the best way to do this, but if you're willing to tolerate me pulling a few rabbits from hats, let's give it a try ----.

Given a chemical reaction

A + B + ... = M + N + ...,

we mean that the reactants A, etc. are in equilibrium with the products M, etc., and that the reaction is reversible (M + ... = A + ...).

The free energy change for the reaction is sum of the free energies of the products minus the sum of that for the reactants as the reaction is written (2 of this + 1 of that +3 of the other reacts to form 1 of something else plus 3 of some other else, or vice versa). Make sense so far?

Free energy is denoted with a bold-face, upper-case, F or G, sometimes italicized (I haven't got to the pt. I can drive the new forum editor quickly enough to avoid being logged off and losing everything), and is more strictly called the "Gibbs free energy."

Now for the rabbit from the hat --- the standard free energy change for the reaction as written is equal to the product of the gas constant, absolute temperature, and the natural log of the equilibrium constant,

del G = RTlnK,

where K is what you're asking about.

If you double the number of moles on each side, you double the free energy, which doubles the log of K, or squares K as you've apparently already figured out.

Still with me? We'll break for questions.
 
  • #6
I "grabbed the start of the string", and will work on it for some time, then will come back for questions.
If you are wondering "Wow! this fast !", well i knew a little about energies ... etc from old times, so i will work on my old info, and what you just told me (G = R*T*ln(K)) to try to understand it.
...
As i am writting this i got just a little question, why the logarthim to the base (e), i mean why is it ln() and not log10 for example ? Any particular reason or it just comes this way ?

Thanks a lot.
 

What is Kc?

Kc is the equilibrium constant, which is a numerical value that represents the ratio of product concentrations to reactant concentrations at equilibrium in a chemical reaction.

How is Kc calculated?

Kc is calculated by taking the concentration of the products raised to their coefficients in the balanced chemical equation, divided by the concentration of the reactants raised to their coefficients.

What does a high value of Kc indicate?

A high value of Kc indicates that at equilibrium, there is a high concentration of products and a low concentration of reactants. This means that the reaction strongly favors the formation of products.

What does a low value of Kc indicate?

A low value of Kc indicates that at equilibrium, there is a low concentration of products and a high concentration of reactants. This means that the reaction strongly favors the formation of reactants.

What factors can affect the value of Kc?

The value of Kc can be affected by changes in temperature, pressure, and the initial concentrations of the reactants and products. Adding a catalyst or changing the volume of the reaction vessel can also alter the value of Kc.

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