Standard Gibbs' energy change

Tags: energy, gibbs, standard
 P: 286 My thermodynamics textbook says that chemical potential can take any of the following symbols: μ (condition-dependent), μ∇ (standard), μΘ (standard) (superscript plimsoll line) and μO (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 ΔHΘ symbol (and ΔSΘ similarly). However, ΔGO is used to denote ΔG under standard conditions of 298.15 K, 1 bar pressure. Should this more correctly be ΔGΘ (so we write ΔGΘ=ΔHΘ-T·ΔSΘ, so the conditions are the same)? Or perhaps they are all actually O? Also, does ΔGΘ 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=ΔGΘ+R·T·loge(Q) and now ΔG=ΔGΘ under such conditions? If not, what is the actual definition of ΔGΘ, and with regards to ΔG itself?
 Sci Advisor P: 3,121 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.
P: 286
 Quote by 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.
I'm getting confused now. How exactly is ΔGr° to be defined? I know that ΔGr°=ΔHr°-T*ΔSr°. 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 ΔGr°=ΔHr°-298.15*ΔSr° 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 ΔGr°=-R*T*loge(K) and now ΔHr°-298.15*ΔSr°=-R*T*loge(K) and our system crashes: T=(ΔHr°-298.15*ΔSr°)/(-R*loge(K)), and far from 298.15 K, our answer will be undefined for K=1.

So the question becomes - what actually is ΔGr° with relation to ΔGr? Evidently it is incorrect to say "standard conditions of 298.15 K and 1 bar pressure" alone because clearly ΔGr° does exhibit temperature dependency due to the T variable in ΔGr°=ΔHr°-T*ΔSr°.

P: 3,121

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
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P: 347
 Quote by Big-Daddy I'm getting confused now. How exactly is ΔGr° to be defined? I know that ΔGr°=ΔHr°-T*ΔSr°. 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 ΔGr°=ΔHr°-298.15*ΔSr° 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 ΔGr°=-R*T*loge(K) and now ΔHr°-298.15*ΔSr°=-R*T*loge(K) and our system crashes: T=(ΔHr°-298.15*ΔSr°)/(-R*loge(K)), and far from 298.15 K, our answer will be undefined for K=1. So the question becomes - what actually is ΔGr° with relation to ΔGr? Evidently it is incorrect to say "standard conditions of 298.15 K and 1 bar pressure" alone because clearly ΔGr° does exhibit temperature dependency due to the T variable in ΔGr°=ΔHr°-T*ΔSr°.
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
 Sci Advisor P: 3,121 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.
PF Patron
P: 347
 Quote by 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.
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

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