Consider the two equations that deal with Delta G.

Equation 1:

Delta G **STANDARD** = – 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 Delta G **STANDARD** can be thought of as a guide to the ratio of the amount of products to the amount of reactants at equilibrium, **and should NOT be thought of as a predictor of the feasibility of the reaction.**

IF it so happens that products and reactants are equally favored at equilibrium, then Delta **G STANDARD** is zero, **BUT Delta G STANDARD is not *necessarily* ZERO at equilibrium (I think this is the key).**

Equation 2:

Delta G = Delta G **STANDARD** + RT ln Q

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

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

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

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

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

IN short, it is Delta G (NOT Delta G **STANDARD**) that will be zero at equilibrium and the sign of IT (in combination with Delta G **STANDARD** and RT ln Q in Equation #2.), will define which way the reaction proceeds.