Dear Momo Lama

Standard condition means the pressure 1 bar and Temp 298K, ΔG° is the measure of Gibbs Free Energy (G) - The energy associated with a chemical reaction that can be used to do work change at 1 bar and 298 K, delta G "naught" (not not) is NOT necessarily a non-zero value. ΔG° = -RT ln(K), So ΔG° = 0, if K = 1. In other words, if product and reactant are equally favoured at equilibrium, it's because there is no difference speaking. ΔG° is always the same for a given reaction. ΔG does depend on your conditions, but still relates to ΔH° and ΔS°. Consider the following:

ΔG = ΔG° + RT ln(P) Where P is the reaction quotient, the ratio of products to reactants at some state. It is equal to K if the system has reached equilibrium. Subbing in ΔG°=ΔH°-TΔS°,

ΔG = ΔH°-TΔS° + RT ln(P)

So, ΔG definitely does relate to those two quantities. ΔH° and ΔS° represent the change in enthalpy and entropy between product and reactant. but they DO NOT mean a "100% complete reaction." They determine the energetic difference (at a given temperature). This difference in energy determines the composition at equilibrium. It has nothing to do with the "completeness" of the reaction, which is a kinetic question. So now in most questions, such as in phase changes questions they would give us the ΔH° and ΔS° and they then would want us to find the temperature at which equilibrium takes place. So using the formula ΔG°=ΔH°-TΔS° we would let ΔG° be 0 and solve for T.  ΔG° is a non-zero value and that we shouldn't be able to use ΔH° or ΔS° to find ΔG because either ΔH° or ΔS° represents 100% complete reaction.

Regards,

Prem Baboo

You should read a good book on Thermodynamics, like MacGlashan's. Standard states are simply reference values that are required as energies are not absolute quantities but are, like velocities (KE =1/2mv^2),only relative ones.