Ice-cream Posted June 15, 2005 Posted June 15, 2005 with the equilibrium constant, when K<1 the reactants are favoured, when K>1 the products are favoured but when K=1...what happens? also, does any1 know whether the terms "carbanion" and "carbocation" refer to a whole molecule or just to the C in the molecule?
Yggdrasil Posted June 16, 2005 Posted June 16, 2005 When K = 1, there will be equal concentrations of products and reactants at equilibirum. In terms of thermodynamics, this corresponds to a reaction with no change in Gibbs free energy ([math]\Delta G = 0[/math]). The terms carbocation and carbanion can refer to both the entire moleculre or the specific carbon.
DQW Posted June 20, 2005 Posted June 20, 2005 with the equilibrium constant, when K<1 the reactants are favoured, when K>1 the products are favoured but when K=1...what happens? What do you mean by "favored" ? At equilibrium, nothing is favored, no matter what the value of K. Also, for most reactions, K is not dimensionless; so changing the units of measurement will change the value of K. So the statement about K being greater than or less than 1 must be made with care. What one can say is the following : In commonly used units (mol/L and powers thereof), if K >> 1, then (the reaction goes nearly to completion before reaching equilibrium, or) the equilibrium concentrations of at least one of the reactants is small compared to their initial concentrations AND when K << 1, the change in concentrations of all the reactants is small compared to the initial/equilibrium concentrations.
gnpatterson Posted June 21, 2005 Posted June 21, 2005 I think I see what you are saying but could I confirm that have got it right by asking if you really mean K<<1? Wouldnt it be more meaningful to say the 1/K >> 1 as K cant go negative and zero doesnt qualify as "very much less than" one?
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