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Kc is the constant of equilibrum 
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#1
Apr903, 10:06 AM

P: 353

K_{c} 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 K_{c}=([ 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 K_{c} will get squared, but this seems wrong since both fomulas are for the same reaction ! Anyone can explain what is happening plz ? Thanks in advance 


#2
Apr903, 11:00 AM

P: 485

That should be a 6D and not a 3D in your second reaction.
I suspect that to be your problem. 


#3
Apr1003, 07:49 PM

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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
Apr1103, 12:10 PM

P: 353

Kc is the constant of equilibrum



#5
Apr1103, 05:28 PM

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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 boldface, uppercase, 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
Apr1203, 01:34 PM

P: 353

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 log_{10} for example ? Any particular reason or it just comes this way ? Thanks a lot. 


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