Re: WHAT EFFECTS equilibrium constant?

From: Boaz (bhaberman_at_softhome.net)
Date: 06/01/04 

Date: 1 Jun 2004 16:37:16 -0700

At equilibrium, G^o = -RT ln K, where G^o is the standard free energy
for the reaction, and K is the equilibrium constant. Since G^o and R
are constant, temperature and ln K are inversely proportional, so K
depends on temperature. However, there is no dependence on pressure.

Using the equilibrium constant, one can demonstrate this aspect of Le
Chatelier's principle, because:

AB(g) -> A(g) + B(g)

([A][B])/([AB])

(n[A]n[B])/(n[AB]) = (n^2[A][B])/([AB])

multiplying pressure by some factor will increase [A][B] by n^2, but
[AB] only increases by a factor of n (think about it). This will make
the equilibrium constant shift to the side with a greater number of
gas molecules, because that side has the greater power. In this case
this is K > Keq, so K must decrease, and the reaction must shift to
the left.

I hope this wasn't too obfuscated.