Confusion about Keq and Standard Conditions

[homiehomes]

I've spent the last hour searching through threads but my question still remains.

Obviously, most reactions will not have Keq=1 at equilibrium-- so why do standard conditions say a system is at equilibrium when concentrations of reactant and products are 1M (or Keq=1?)

I feel like there is something simple that I confused myself with Thanks for any help!!

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[desperadoflakes]

No, Keq won't be = 1 at equilibrium (unless the ∆G' of a reaction is 0, and the reaction is totally reversible).

Where do you find that that's what standard conditions say?

ΔG°'=-RTlnKeq. If you look in a biochemistry textbook or reference almanac, you'll find a huge table of ΔG°' values for a bunch of reactions. If the values are below 0, then the reactions are spontaneous at biochemical standard conditions.

ΔG° is regular standard conditions, and ΔG°' is used in biochemistry. The regular standard conditions are that the reaction takes place at 298T, 1 ATM, and that the initial concentrations of all reactant and product species are 1M. This means that we are assuming that Q = 1, but NOT Keq. The difference with ΔG°' (biochemistry standard conditions) is that we also add in the condition that pH = 7 (so [H+] = 10^-7), and also in both conditions if you want to be technical than [H2O] = 55.5M, but it doesn't matter because H2O doesn't factor into our Q or Keq equations being a pure liquid.

In the equation ΔG°'=-RTlnKeq , you're not plugging in those 1M values anywhere. If you look in an textbook and you find that the ΔG°' of a certain reaction is -1000J/mol, what that means is that when you start with 1M of all your reactants and your products (or 1E-7 for H+ or OH-) at 298K, your Keq will be equal to e^((-1000/(-8.314*298)) = 1.5

So in this case, your favorable reaction will make it so you end up with Keq = 1.5, which means that half of your initial reactants will go to the right, so now you have 0.5M of reactants and 1.5M of products. We can see this mathematically by looking at the extended equation.

∆G = ΔG°' + RTlogQ
This reaction allows us to take our ΔG°' which tells us what our reaction looks like at standard conditions, and find the REAL ∆G using OUR conditions. So if we start off with 1M of all reactants and products, then Q will be 1, and logQ will be 0, and then
∆G = ΔG°' = -RTlnKeq

The other way to simplify this is by simplifying that equation.

∆G = ΔG°' + RTlogQ ; ΔG°' = -RTlnKeq
∆G = -RTlnKeq + RTlogQ
∆G = -RT(lnKeq - logQ)

So it turns out that ΔG°' = -RTlnKeq, BUT ∆G = -RTln(Keq - logQ). Under standard conditions, Q = 1, so ∆G = ΔG°'

hope that makes sense.

 

[homiehomes]

Your explanation is crystal clear, thank you so much !!