delta G and Keq

[sv3]

Hi there,

I'm confused about the following relationship:
delta G <0, Keq>1
delta G = 0, Keq = 1,
delta G>1, Keq<1

when does this hold true and when does it not? I thought i understood this when doing chemistry, but now that I'm onto bio, a section stated that when delta G = 0, equilibrium was established - this was even though Keq was 1000. This actually made sense to me (the Keq for a reaction at equilibrium can be any number afterall), and has caused me to question the above relationship I've seen in some prep books where Keq has to be 1 when delta G = 0. I obviously don't understand it as well as I thought so was hoping someone could clairfy for me. This seems to be an important relationship to understand, hence my post

thanks in advance
steve

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[Dr Gerrard]

aA + bB --> cC + dD

keq = ([A]^a * ^b)/([C]^c * [D]^d)

That is true for all cases.

Now, if delta G is less than 0, that means the reaction is spontaneous.

This is true for all cases.

If I recall correctly, delta G does not have to be zero in order for the reaction to be in equilibrium.

That means the above reaction will proceed to the right spontaneously, and at equilibrium, the concentration of the products will be greater than the concentration of the reactants.

According to the Keq formula above, if the concentration of the products are greater than the concentration of the reactants, than the nominator is greater than the denominator and thus Keq is greater than one.

 

[sv3]

aA + bB --> cC + dD

keq = ([A]^a * ^b)/([C]^c * [D]^d)

That is true for all cases.

Now, if delta G is less than 0, that means the reaction is spontaneous.

This is true for all cases.

If I recall correctly, delta G does not have to be zero in order for the reaction to be in equilibrium.

That means the above reaction will proceed to the right spontaneously, and at equilibrium, the concentration of the products will be greater than the concentration of the reactants.

According to the Keq formula above, if the concentration of the products are greater than the concentration of the reactants, than the nominator is greater than the denominator and thus Keq is greater than one.

hi,

think your eq formula is backwards - should be products over reactants no?

Its the delta G = Standard G +RTlnK equation that i was having issues with. I think i might be mixing up Keq and K. My initial post had that relationship but I've now double checked and realize my two sources use the same relationships, but one does it with reference to K and delta G, and one does it for Keq and delta G. I'm thinking one of them is mistaken. I'll just have to keep playing with this to figure it out i guess.

 

[Dr Gerrard]

Yes, I am sorry, I was incorrect before.

I think the following is correct.

At equilibrium, delta G is equal to zero.

Thus, you have this equation: delta G = Standard G +RTlnK

If K is less than one, then the second term is negative, thus standard G has to be greater than zero.

If K is equal to one, then standard G has to equal 0.

If K is greater than one, then standard G has to be less than zero.

K does not HAVE to equal anything to be at equilibrium, because K is a constant that tells you the concentrations of products and reactants at equilibrium and is dependent on temperature.

These relationships only hold true when delta G is equal to 0 and the reaction is at equilibrium.

 

[sv3]

Yes, I am sorry, I was incorrect before.

I think the following is correct.

At equilibrium, delta G is equal to zero.

Thus, you have this equation: delta G = Standard G +RTlnK

If K is less than one, then the second term is negative, thus standard G has to be greater than zero.

If K is equal to one, then standard G has to equal 0.

If K is greater than one, then standard G has to be less than zero.

K does not HAVE to equal anything to be at equilibrium, because K is a constant that tells you the concentrations of products and reactants at equilibrium and is dependent on temperature.

These relationships only hold true when delta G is equal to 0 and the reaction is at equilibrium.

Actually K isn't a constant. Its like Keq but can be concentrations at any point in time so therefore its not a constant. What i bolded above is the description for Keq. I think your math regarding standard G is still correct however. This is really killing me now! thanks for the reply though

Also, you first state if delta G is 0, then its at equilibirium. But you finish by saying there's 2 conditions.....so i can't which is which? If delta G = 0, then do you need to know anything else to know the reaction is at equilibrium?

sv3

 

[Rach83]

hi,

think your eq formula is backwards - should be products over reactants no?

Its the delta G = Standard G +RTlnK equation that i was having issues with. I think i might be mixing up Keq and K. My initial post had that relationship but I've now double checked and realize my two sources use the same relationships, but one does it with reference to K and delta G, and one does it for Keq and delta G. I'm thinking one of them is mistaken. I'll just have to keep playing with this to figure it out i guess.

Well most sources I have seen use Keq (or K) for the equilibrium, and Q for when things are still changing. So it might be that both of your sources are talking about the same equilibrium K. I have only ever seen this equation like this:
G=Stand. G + RT ln Q.

Hope this actually helps and that I'm understanding your problem. 🙂

 

[Dr Gerrard]

Actually K isn't a constant. Its like Keq but can be concentrations at any point in time so therefore its not a constant. What i bolded above is the description for Keq. I think your math regarding standard G is still correct however. This is really killing me now! thanks for the reply though

Also, you first state if delta G is 0, then its at equilibirium. But you finish by saying there's 2 conditions.....so i can't which is which? If delta G = 0, then do you need to know anything else to know the reaction is at equilibrium?

sv3

I guess I am learning some of this as well.

To your first point though. If K isn't constant, why is it called the equilibrium constant? I think the K in the equation we are using is meant to substitute for the equilibrium constant, because the K can only go inside the natural log as a substitute for Q.

Second, if delta G = 0, the reaction is at equilibrium. That will always be true.