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Standard Gibbs' energy change

by Big-Daddy
Tags: energy, gibbs, standard
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Big-Daddy
#1
Mar29-13, 11:07 AM
P: 343
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-TΔSΘ, 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+RTloge(Q) and now ΔG=ΔGΘ under such conditions? If not, what is the actual definition of ΔGΘ, and with regards to ΔG itself?
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DrDu
#2
Mar29-13, 01:34 PM
Sci Advisor
P: 3,564
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
#3
Mar29-13, 04:20 PM
P: 343
Quote Quote by DrDu View Post
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.

DrDu
#4
Mar30-13, 07:36 AM
Sci Advisor
P: 3,564
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
morrobay
#5
Apr11-13, 06:26 AM
P: 376
Quote Quote by Big-Daddy View Post
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
DrDu
#6
Apr11-13, 11:14 AM
Sci Advisor
P: 3,564
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
#7
Apr12-13, 04:20 AM
P: 376
Quote Quote by DrDu View Post
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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