
#1
Mar2913, 11:07 AM

P: 335

My thermodynamics textbook says that chemical potential can take any of the following symbols: μ (conditiondependent), 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? 



#2
Mar2913, 01:34 PM

Sci Advisor
P: 3,370

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. 



#3
Mar2913, 04:20 PM

P: 335

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}°. 



#4
Mar3013, 07:36 AM

Sci Advisor
P: 3,370

Standard Gibbs' energy change
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 



#5
Apr1113, 06:26 AM

PF Gold
P: 362

in their standard states. Where ΔG = ΔHTΔS 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 



#6
Apr1113, 11:14 AM

Sci Advisor
P: 3,370

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.




#7
Apr1213, 04:20 AM

PF Gold
P: 362

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 


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