Consider the two** equations that deal with Delta G (∆G).

***Since this post was originally written in January 2012, the AP exam has changed. One of the changes was to remove equation #2 below from the equations & constants sheet. As such, I think that knowledge of it, and the consequences associated with it, are unlikely to be tested quantitatively on the exam in the future, but nevertheless, I still feel that understanding that conditions other than standard ones will cause ∆G to take on new values is a useful reference point.*

Equation 1:

∆G° = – RT ln K

Since K is the equilibrium constant, we *are* at equilibrium, the amounts of products and reactants in the mixture are fixed, and the sign of ∆G**°** can be thought of as a guide to the ratio of the amount of products to the amount of reactants at equilibrium and therefore the *thermodynamic favorability* of the reaction.

If it so happens that products and reactants are equally favored at equilibrium, then ∆G° is zero, **BUT ∆G° is not *necessarily* ZERO at equilibrium.
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Equation 2:

∆G = ∆G° + RT ln Q

Since Q is NOT the K, and we are NOT necessarily at the equilibrium position, the sign of ∆G can be thought of as a predictor about which way the reaction (that has reactants and products defined by Q), will go.

If ∆G° is negative at equilibrium, then we will have lots of products at equilibrium, meaning Q needs to be bigger (greater than 1) to approach K. As Q gets larger (i.e., as we get more products), the term ‘RT ln Q’ gets increasingly positive, and eventually adding that term to a negative ∆G°**, **will make ∆G = 0, equilibrium will be established and no further change occurs.

It is possible that Q could already be too large and therefore ∆G is positive. IF so, then the reaction will need to from more reactants, reduce the value of Q, and allow ∆G to reach zero, i.e., allow equilibrium to be established.

If ∆G° is positive at equilibrium, then we will have lots of reactants at equilibrium, meaning Q needs to be smaller (less than 1) to approach K. As Q gets smaller (i.e., as we get more reactants), the term ‘RT ln Q’ gets increasingly negative, and eventually adding that term to a positive ∆G°**, **will make ∆G = 0, equilibrium will be established and no further change occurs.

It is possible that Q could already be too small and therefore ∆G is negative, IF so, then the reaction will need more products, increase the value of Q, and allow ∆G to reach zero, i.e., allow equilibrium to be established.

In short, it is ∆G (NOT ∆G°) that will be zero at equilibrium and the sign of *it* (generated by the combination of ∆G° and RT ln Q in Equation #2), will define which way the reaction proceeds.

But delta G naught is the delta G at standard condition. 25 C, 1 atm. So delta G naught is constant for a given reaction. It is related to K at the equilibrium temp since then delta G is 0.

The deviation of delta G from delta G0 is given by: delta G = delta G0 + RTlnQ, where Q = product/reactants expression.

When Equilibrium is obtained, delta G = 0, So delta G0 = -RTln K, ie. Q is now K as Q was for non equilibrium.

So delta G0 is not necessarily 0 be it at 25 deg C and 1 atmosphere or otherwise. Delta G = 0 at equilibrium, not delta G0.