Relating Gibbs free energy and Equilibrium constant

[vegan1]

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Could someone kindly explain to me why deltaG and Keq relate in the following manner (according to EK Chemistry Electrochemistry chapter - p. 118):

if K=1 then deltaG(naught) = 0

if K>1 then deltaG(naught) < 0

if K<1 then deltaG(naught) > 0

I was thinking that if Keq>1, then [products]>[reactants] and the rxn should proceed backwards, which in my mind, should indicate a positive deltaG(naught). Where am I wrong here?

vegan1

 

[rom73085]

Well,
If Q<K then K <1 and thus - delta G which is spontaneous indicating too many products.

Likewise, if Q>K then K>1 thus + delta G and non spontaneous indicating too many reactants.

You are right to say that if K>1 then the rxn would indicate a positive change in free energy, but this would be indicating that products < reactants.

Make sense?

 

[GRod18]

Do you know the equation Delta G (naught) = -RT ln keq ? this equation is a must for the MCAT.

As you can see here if you take the natural log of a number less than 1 (you would get a negative number, but a double negative would = a positive Delta G (naught), indicating that the reaction is NOT spontaneous in the forward direction and that the reactants are favored at equlibirum. (you should know that - delta G = spontaneous in the forward direction and that the products are favored in this case)

Delta G is always equal to zero at equalibrium so at Keq of 1 (plugged into equation) you can see that Delta G = 0. Meaning either reactant or product is favored.

 

[fizzgig]

i dont know if this is right.

if k>1, then AT EQLB you will have more product. so the forward reaction is the favored reaction, and standard free E is negative.

if k<1 then at eqlb you will have less product vs reactants. so the reverse reaction is the favored reaction, and standard free E is positive.

if k=1 then at eqlb you have equal products and reactants, so both reactions are equally favored.

whether the actual reaction will run forwards or backwards depends on ACTUAL concentrations of prod/react of course, so if k>1 you 'want' more products at eqlb, but you can still overshoot that and get waaay too much product, in which case the deltaG (not dG naught) would still be positive if you removed the stimulus you used to push it past eqlb (like raising the T or something).

if this thinking is totally off someone let me know and i'll edit to not confuse people.

 

[vegan1]

if k>1, then AT EQLB you will have more product. so the forward reaction is the favored reaction, and standard free E is negative.

if k<1 then at eqlb you will have less product vs reactants. so the reverse reaction is the favored reaction, and standard free E is positive.

if k=1 then at eqlb you have equal products and reactants, so both reactions are equally favored.


This makes sense to me.

 

[bullsfan1986]

I think if you use a potential energy (disregard KE) analogy, free energy and k become a lot more intuitive.

so if k=0 there is not going to be any rxn and thus deltaG=0

if k>1 (fast) PE is maximal (at the top of the hill) and the rxn will go downhill (path of least resistance) p>r

if k<1 (slow) PE is minimal (not zero) and thus getting up the hill will take the path of most resistance thus r>p (rev rxn will be favored)

One of my gen chem teachers explained it like this and it really clicked.

 

[bullsfan1986]

and by "no reaction" I mean r and p will be in equilibrium.