What Is the Correct Symbol for Gibbs' Energy Under Standard Conditions?

[Big-Daddy]

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?

 

[DrDu]

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.

 

[Big-Daddy]

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°.

 

[DrDu]

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

 

[morrobay]

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°.

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

 

[DrDu]

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.

 

[morrobay]

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