My thermodynamics textbook says that chemical potential can take any of the following symbols: μ (condition-dependent), μ^{∇} (standard), μ^{Θ} (standard) (superscript plimsoll line) and μ^{O} (standard), with which one of the latter 3 being used depending on the temperature chosen to define standard state. I have seen that the notation typically used for ΔH under standard conditions of 298.15 K, 1 bar pressure is the ΔH^{Θ} symbol (and ΔS^{Θ} similarly). However, ΔG^{O} is used to denote ΔG under standard conditions of 298.15 K, 1 bar pressure. Should this more correctly be ΔG^{Θ} (so we write ΔG^{Θ}=ΔH^{Θ}-T·ΔS^{Θ}, so the conditions are the same)? Or perhaps they are all actually ^{O}? Also, does ΔG^{Θ} for a reaction actually refer to the value of ΔG for that reaction under standard conditions of 298.15 K, 1 bar pressure? If so, wouldn't that mean that the equilibrium constant must be 1 under standard conditions (which is not necessarily true), as a result of ΔG=ΔG^{Θ}+R·T·log_{e}(Q) and now ΔG=ΔG^{Θ} under such conditions? If not, what is the actual definition of ΔG^{Θ}, and with regards to ΔG itself?
To your second question. Yes, K=1 at the standard state, but this is not an equilibrium situation and the standard state is often only a theoretical construct which is impossible to realize experimentally. E.g. you can't have a mixture of hydrogen and oxygen with partial pressure 1 bar of each component and total pressure being 1 bar, too.
I'm getting confused now. How exactly is ΔG_{r}° to be defined? I know that ΔG_{r}°=ΔH_{r}°-T*ΔS_{r}°. But doesn't the ° symbol mean "standard conditions of 298.15 K and 1 bar pressure" or not - because if so then we may as well write ΔG_{r}°=ΔH_{r}°-298.15*ΔS_{r}° and I'm sure this is wrong, because now let's look for the temperature at which K=1 (which should, according to what I showed in my OP, be 298.15 K, but is as I will show now indeterminate) through ΔG_{r}°=-R*T*log_{e}(K) and now ΔH_{r}°-298.15*ΔS_{r}°=-R*T*log_{e}(K) and our system crashes: T=(ΔH_{r}°-298.15*ΔS_{r}°)/(-R*log_{e}(K)), and far from 298.15 K, our answer will be undefined for K=1. So the question becomes - what actually is ΔG_{r}° with relation to ΔG_{r}? Evidently it is incorrect to say "standard conditions of 298.15 K and 1 bar pressure" alone because clearly ΔG_{r}° does exhibit temperature dependency due to the T variable in ΔG_{r}°=ΔH_{r}°-T*ΔS_{r}°.
I would have to look up the details myself. A very good reference on standard states is: Klotz, Irving M. / Rosenberg, Robert M. Chemical Thermodynamics Basic Concepts and Methods
ΔG° = ΔH°-TΔS° This is the free energy change with standard conditions @ 298 K from elements in their standard states. Where ΔG = ΔH-TΔS is for calculating when ΔG becomes negative with increasing temperature. Also regarding solving for T when K =1 , its my understanding that its when ΔG = 0 that K =1
If K=1 the logarithm vanishes and ##\Delta G=\Delta G^0##. K=1 in general does not correspond to an equilibrium situation (even less at fixed temperature so ##\Delta G## won't be 0. In contrast, ##\Delta G^0=-RT \ln K## holds only in equilibrium.
In post # 3 Big-Daddy is solving for T with ΔG° = ΔH° - TΔS° and ΔG°(298K) = -RTlnK Standard free energy changes @ 298K. And in ΔG° = -RTlnK 'K' can have different values and when K = 1 , ΔG° = 0 Not questioning your last post - just for clarification