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Offline Yggdrasil

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Re: Gibbs' Free energy Olympiad Question
« Reply #15 on: March 29, 2013, 11:14:41 AM »
Here, it doesn't make much sense to talk about an equilibrium constant (Kc) since both reactants and products are solids.

Offline Corribus

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Re: Gibbs' Free energy Olympiad Question
« Reply #16 on: March 29, 2013, 11:36:38 AM »
Well it's not that it doesn't make sense.  It's just that it's equal to 1 because the activities of products and reactants are both unity.
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Offline Big-Daddy

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Re: Gibbs' Free energy Olympiad Question
« Reply #17 on: March 29, 2013, 11:56:46 AM »
Well it's not that it doesn't make sense.  It's just that it's equal to 1 because the activities of products and reactants are both unity.

Ah but then doesn't that mistakenly suggest that equilibrium will not be reached at all except at T=286.4 K?

In any case this is a secondary issue for me at the moment; the first is my understanding of ΔG°!

Offline Corribus

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Re: Gibbs' Free energy Olympiad Question
« Reply #18 on: March 29, 2013, 12:36:21 PM »
No, sorry, nevermind, that was wrong.  The acitivities of both are not unity, because the gray tin is not considered standard state.  Equilibrium constant is still a useful concept, even for solids, but chemical activities have to be used because concentrations are meaningless.  We usually don't include pure solids and liquids in equilibrium expression because they are considered to be in their standards states and have specified chemical activities equal to one.  Interconversion of two states of a solid would require one to not be in the standard state (obviously), but honestly I'd have to look into how to determine a priori what the activity is of a solid in another state because I don't know.  If I figure it out, I'll let you know.  Sorry, doing manipulations of solid-solid equilibria isn't something I've a lot of experience with. Learning as I go. :)

What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

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Re: Gibbs' Free energy Olympiad Question
« Reply #19 on: March 29, 2013, 12:54:26 PM »
No, sorry, nevermind, that was wrong.  The acitivities of both are not unity, because the gray tin is not considered standard state.  Equilibrium constant is still a useful concept, even for solids, but chemical activities have to be used because concentrations are meaningless.  We usually don't include pure solids and liquids in equilibrium expression because they are considered to be in their standards states and have specified chemical activities equal to one.  Interconversion of two states of a solid would require one to not be in the standard state (obviously), but honestly I'd have to look into how to determine a priori what the activity is of a solid in another state because I don't know.  If I figure it out, I'll let you know.  Sorry, doing manipulations of solid-solid equilibria isn't something I've a lot of experience with. Learning as I go. :)

Can you read my last post on ΔGr°? That's a much more fundamental problem of mine. Not just applying to solid-solid equilibria but to the maths of equilibrium in general.

Offline Corribus

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Re: Gibbs' Free energy Olympiad Question
« Reply #20 on: March 29, 2013, 08:19:43 PM »
This is the way I see it.  Please excuse if I’m writing stuff you already know.  I am trying to be as clear as possible, both for myself and for you.
 
Say you have a substance A that converts to B.  Now if you mix A and B together at a certain temperature T, what happens?
 
A ::equil:: B
 
If you start with all A, you know you'll begin to form some B after some time passes.  If you start with all B, you know you'll start to form some A.  At  equilibrium (infinite time), you know the relative concentrations (ratio) of A and B will be the same regardless of whether you start with all A and all B.  This ratio is specified by the equilibrium constant, K.  And K will be dependent on the temperature.
 
We introduce the concept of ΔG°, the Gibbs energy of formation.  This is equal to the difference between the standard Gibbs energy of formation of B and the standard Gibbs energy of formation of A.  In turn, the  standard Gibbs energy of formation of each substance is the amount of free energy gained or lost from the formation of that substance at temperature T.  These values are in turn expressed relative to the standard form of the  substance.  And they will be dependent on T because the amount of energy required to form a substance at a particular temperature is obviously related to T.  Thermodynamic values are almost always expressed relative to other values. 

Another way we can calculate ΔG° is from the equation
 
ΔG° = ΔH° – TΔS°
 
The change in the Gibbs energy of formation is temperature dependent, but most of the time it is expressed at 298.15 K because this is usually the value at which ΔH° and TΔS° are measured.  (Although, ΔH° and TΔS° are usually taken as temperature independent quantities, so in practice it doesn’t matter, let’s be pedantic here and say it does.)
 
Now again, at temperature T if we start with concentrations of A and B at exactly equal to the ratio specified by the equilibrium constant at that temperature, then we are already at equilibrium at that temperature and we will have no net conversion of A to B or B to A.  (We will see in a minute this is the case where Δ G = 0 because there is no net driving force in either direction.)
 
But let’s say we start with an A to B ratio (at temperature T) that is LARGER than the ratio specified by the equilibrium constant.  In this case Q, the reaction quotient, is smaller than K (Q = [ B]/[A]).  We know by intuition at this point that if this happens, we will get a spontaneous net conversion of some of the A into B, and this will continue until A and B are at the magic ratio specified by the equilibrium constant.  The farther away the relative concentrations of A and B start from this equilibrium ratio, the greater this driving force for conversion will be.  Also, if we start with an A to B ratio that is SMALLER than the ratio specified by the equilibrium constant, Q is larger than K and we will get a spontaneous net conversion of some of the B into A, and again this will continue until we reach the equilibrium ratio of [A] to [ B], specified by K, at temperature T.  The equilibrium point will be the same regardless of the starting point!
 
Let’s introduce ΔG now.  We’ve said that depending on the real ratio of A to B, with respect to the equilibrium ratio A to B, the reaction will tend to move one way or the other spontaneously. The direction of this movement, and the magnitude of the driving force, is embodied in ΔG .  ΔG is dependent on the instantaneous state of the system at time t, where the concentrations of A and B are [A] and [ B], and the temperature is T.   When the [A] and [ B] are far from the equilibrium value, the magnitude of ΔG will be large; when [A] and [ B] are at the equilibrium ratio, the magnitude of ΔG is zero.
 
We know that
 
ΔG = ΔG° + RT ln Q
 
As we’ve said, ΔG is a measure how spontaneous the consumption of A to B is (or B to A, depending on the sign of ΔG ) and is equal to zero at equilibrium.  In a lot of textbooks on this subject they present a hypothetical plot of ΔG as a function of the mole fraction of one of the substances A or B (for a given T), which has a parabolic kind of shape.  Where the minimum of this plot would be in relation to pure A or pure B is related to ΔG°.  So ΔG° is a description of how how far to the left or right a reaction is at equilibrium.  (As an example, if K = 1, then the concentration of A and B are equal at equilibrium, ΔG°is zero and the minimum of this hypothetical plot of ΔG vs. Q would be at a value of 1, because Q = K at equilibrium.

At equilibrium, then, ΔG = 0 and Q = K, so ΔG° = - RT ln K.  In the case that K = 1, ΔG° = 0 because the equilibrium concentrations of A and B favor neither A or B (there are equal amounts of both).  In this case ΔG for any given concentration of A and B just is = RT ln Q.  What this essentially means is that if [ B] = 2[A], the magnitude of ΔG is the same as if [A] = 2[ B] (but opposite sign).  This makes perfect sense because the driving force should be the same, but in opposite directions, along the chemical potential surface if the concentrations of A and B are switched - the equilibrium always will drive to a 50:50 mixture.

The final thing to consider is the effect of temperature.  Temperature changes things only by shifting the equilibrium.  The ΔG° value will also change if it is determined at a different temperature.  In principle, you would need to determine a new ΔG° by either looking of Gibbs energies of formation at the different temperatures or by looking up enthalpies and entropies of formation at the different temperatures.  The Gibbs energies of formation specify the free energy change for forming a given substance relative to the standard state of that substance, and this will depend on the temperature.  In practice, however, it is almost always assumed that ΔH° and ΔS° are temperature independent, so what you have to do most of the time to determine ΔG° at a new temperature is to just use the same values of ΔH° and ΔS° and use the new temperature in the familiar equation above.  You can then use this ΔG° value to determine a new equilibrium constant (- RT ln K) at this temperature and all the ΔG values for any given reaction quotient.

However the typical way to measure equilibrium constants at a new temperature is to just use the van't Hoff equation.  (This is what your manipulation in the previous post was driving at, but you were trying to compare two temperature values with only one equilibrium constant.) When you do this you will see all that you really need is ΔH°, which as usual we've assumed to be temperature independent.

ln (K2/K1) = -(ΔH°/R)(T2-1 - T1-1)

Using this you can calculate K at any temperature as long as ΔH° is known at K is known at one temperature (typically at 25 °C, which is the easiest to calculate from tabulated data).  Of course, the accuracy of this approach dependence on the variability of the enthalpy and entropy change wrt to temperature.  Otherwise you need to do it the long way.

In the present problem, equilibrium works the same way, but you can't use concentrations for the equilibrium constants.  You'll have to find some other way to express activity.  As I said, I'm not sure of the appropriate way to do that. 

Hopefully that is helpful and less confusing that previous discussion?  Given my track record lately I'm sure there will be some kind of error in there somewhere that I'll have to later apologize for, but here's keeping my fingers crossed. ;)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

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Re: Gibbs' Free energy Olympiad Question
« Reply #21 on: March 29, 2013, 09:57:11 PM »
This is the way I see it.  Please excuse if I’m writing stuff you already know.  I am trying to be as clear as possible, both for myself and for you.
 
Say you have a substance A that converts to B.  Now if you mix A and B together at a certain temperature T, what happens?
 
A ::equil:: B
 
If you start with all A, you know you'll begin to form some B after some time passes.  If you start with all B, you know you'll start to form some A.  At  equilibrium (infinite time), you know the relative concentrations (ratio) of A and B will be the same regardless of whether you start with all A and all B.  This ratio is specified by the equilibrium constant, K.  And K will be dependent on the temperature.
 
We introduce the concept of ΔG°, the Gibbs energy of formation.  This is equal to the difference between the standard Gibbs energy of formation of B and the standard Gibbs energy of formation of A.  In turn, the  standard Gibbs energy of formation of each substance is the amount of free energy gained or lost from the formation of that substance at temperature T.  These values are in turn expressed relative to the standard form of the  substance.  And they will be dependent on T because the amount of energy required to form a substance at a particular temperature is obviously related to T.  Thermodynamic values are almost always expressed relative to other values. 

Another way we can calculate ΔG° is from the equation
 
ΔG° = ΔH° – TΔS°
 
The change in the Gibbs energy of formation is temperature dependent, but most of the time it is expressed at 298.15 K because this is usually the value at which ΔH° and TΔS° are measured.  (Although, ΔH° and TΔS° are usually taken as temperature independent quantities, so in practice it doesn’t matter, let’s be pedantic here and say it does.)
 
Now again, at temperature T if we start with concentrations of A and B at exactly equal to the ratio specified by the equilibrium constant at that temperature, then we are already at equilibrium at that temperature and we will have no net conversion of A to B or B to A.  (We will see in a minute this is the case where Δ G = 0 because there is no net driving force in either direction.)
 
But let’s say we start with an A to B ratio (at temperature T) that is LARGER than the ratio specified by the equilibrium constant.  In this case Q, the reaction quotient, is smaller than K (Q = [ B]/[A]).  We know by intuition at this point that if this happens, we will get a spontaneous net conversion of some of the A into B, and this will continue until A and B are at the magic ratio specified by the equilibrium constant.  The farther away the relative concentrations of A and B start from this equilibrium ratio, the greater this driving force for conversion will be.  Also, if we start with an A to B ratio that is SMALLER than the ratio specified by the equilibrium constant, Q is larger than K and we will get a spontaneous net conversion of some of the B into A, and again this will continue until we reach the equilibrium ratio of [A] to [ B], specified by K, at temperature T.  The equilibrium point will be the same regardless of the starting point!
 
Let’s introduce ΔG now.  We’ve said that depending on the real ratio of A to B, with respect to the equilibrium ratio A to B, the reaction will tend to move one way or the other spontaneously. The direction of this movement, and the magnitude of the driving force, is embodied in ΔG .  ΔG is dependent on the instantaneous state of the system at time t, where the concentrations of A and B are [A] and [ B], and the temperature is T.   When the [A] and [ B] are far from the equilibrium value, the magnitude of ΔG will be large; when [A] and [ B] are at the equilibrium ratio, the magnitude of ΔG is zero.
 
We know that
 
ΔG = ΔG° + RT ln Q
 
As we’ve said, ΔG is a measure how spontaneous the consumption of A to B is (or B to A, depending on the sign of ΔG ) and is equal to zero at equilibrium.  In a lot of textbooks on this subject they present a hypothetical plot of ΔG as a function of the mole fraction of one of the substances A or B (for a given T), which has a parabolic kind of shape.  Where the minimum of this plot would be in relation to pure A or pure B is related to ΔG°.  So ΔG° is a description of how how far to the left or right a reaction is at equilibrium.  (As an example, if K = 1, then the concentration of A and B are equal at equilibrium, ΔG°is zero and the minimum of this hypothetical plot of ΔG vs. Q would be at a value of 1, because Q = K at equilibrium.

At equilibrium, then, ΔG = 0 and Q = K, so ΔG° = - RT ln K.  In the case that K = 1, ΔG° = 0 because the equilibrium concentrations of A and B favor neither A or B (there are equal amounts of both).  In this case ΔG for any given concentration of A and B just is = RT ln Q.  What this essentially means is that if [ B] = 2[A], the magnitude of ΔG is the same as if [A] = 2[ B] (but opposite sign).  This makes perfect sense because the driving force should be the same, but in opposite directions, along the chemical potential surface if the concentrations of A and B are switched - the equilibrium always will drive to a 50:50 mixture.

The final thing to consider is the effect of temperature.  Temperature changes things only by shifting the equilibrium.  The ΔG° value will also change if it is determined at a different temperature.  In principle, you would need to determine a new ΔG° by either looking of Gibbs energies of formation at the different temperatures or by looking up enthalpies and entropies of formation at the different temperatures.  The Gibbs energies of formation specify the free energy change for forming a given substance relative to the standard state of that substance, and this will depend on the temperature.  In practice, however, it is almost always assumed that ΔH° and ΔS° are temperature independent, so what you have to do most of the time to determine ΔG° at a new temperature is to just use the same values of ΔH° and ΔS° and use the new temperature in the familiar equation above.  You can then use this ΔG° value to determine a new equilibrium constant (- RT ln K) at this temperature and all the ΔG values for any given reaction quotient.

However the typical way to measure equilibrium constants at a new temperature is to just use the van't Hoff equation.  (This is what your manipulation in the previous post was driving at, but you were trying to compare two temperature values with only one equilibrium constant.) When you do this you will see all that you really need is ΔH°, which as usual we've assumed to be temperature independent.

ln (K2/K1) = -(ΔH°/R)(T2-1 - T1-1)

Using this you can calculate K at any temperature as long as ΔH° is known at K is known at one temperature (typically at 25 °C, which is the easiest to calculate from tabulated data).  Of course, the accuracy of this approach dependence on the variability of the enthalpy and entropy change wrt to temperature.  Otherwise you need to do it the long way.

In the present problem, equilibrium works the same way, but you can't use concentrations for the equilibrium constants.  You'll have to find some other way to express activity.  As I said, I'm not sure of the appropriate way to do that. 

Hopefully that is helpful and less confusing that previous discussion?  Given my track record lately I'm sure there will be some kind of error in there somewhere that I'll have to later apologize for, but here's keeping my fingers crossed. ;)

No I think I may finally have understood! :)
So ΔG° actually is temperature dependent (I'm guessing the ° symbol then only refers strictly to pressure?), but since ΔH° and ΔS° are (approximated to be) temperature independent, then whatever T we want ΔG° for, we can just take the 298.15 K values of ΔH° and ΔS° (under the assumption that temperature modification is negligible) and then plug that in together with the T to ΔG°=ΔH°-T·ΔS°.

Meanwhile, there are no restrictions on conditions in the case ΔG=ΔH-T·ΔS, so we would plan to use temperature-dependent ΔH and ΔS values to reach the temperature dependent ΔG.

However, this would suggest that, if we take the approximation that ΔH and ΔS are not temperature dependent and we are working at 1 bar pressure, we could also have ΔG=ΔH°-T·ΔS° - which would suddenly mean ΔG=ΔG° and this is far too great a generalization to be true when all we've said is that ΔH and ΔS are not temperature dependent (and we're at 1 bar pressure). Particularly as it means K=1 in all such cases, which obviously does not have to be true. So this problem persists!

Thank you for the extra clarity on ΔG itself and driving the reaction a certain direction. I will probably have to return to that shortly in my studies. But for now my understanding of the ΔG/ΔG° difference is still patchy.

Offline Corribus

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Re: Gibbs' Free energy Olympiad Question
« Reply #22 on: March 30, 2013, 12:54:45 AM »
So ΔG° actually is temperature dependent (I'm guessing the ° symbol then only refers strictly to pressure?), but since ΔH° and ΔS° are (approximated to be) temperature independent, then whatever T we want ΔG° for, we can just take the 298.15 K values of ΔH° and ΔS° (under the assumption that temperature modification is negligible) and then plug that in together with the T to ΔG°=ΔH°-T·ΔS°.
Yes that's pretty much it.  Forgetting the Δ for a moment, the ° in a G° indicates that it is the amount of free energy change resulting from the formation of a substance from the standard state of the component elements.  ΔG° for a reaction would effectively be the change in free energy occuring if the reactants in their standard states were converted into products in their standard states.  Pretty much every tabulated value you're going to find is going to be measured at 298.15.

Quote
However, this would suggest that, if we take the approximation that ΔH and ΔS are not temperature dependent and we are working at 1 bar pressure, we could also have ΔG=ΔH°-T·ΔS° - which would suddenly mean ΔG=ΔG° and this is far too great a generalization to be true when all we've said is that ΔH and ΔS are not temperature dependent (and we're at 1 bar pressure). Particularly as it means K=1 in all such cases, which obviously does not have to be true. So this problem persists!
No, here you are confusing the standard enthalpy/entropy changes, which are benchmark values, and the actual enthalpy/entropy changes that occur for a reaction given specific concentrations of reactants and products.  (And therefore you are also confusing ΔG and ΔG°.)

Think of it this way: For a reaction, the standard thermodynamic values represent the change in the thermodynamic values that would result when taking 1 mol of the reactants and converting to one mol of the products at the specified temperature.  This is why they are standard - they are values determined under a specific set of conditions that serve as reference values, because most thermodynamic quantities only have relevance when measured respective to something else.  It is change we are interested in, not absolute quantities.  However most chemical reactions in reality do not relate converting 1 mol of this exactly into one mol of that.  First, there are always equilibria - so conversion isn't complete - and you're likely to have concentrations or amounts of reacting substances that differ from 1 mol.  The DG, DH and DS for a reaction are the actual changes in those values that occur based on the specified conditions.  Those valuese are reported relative to the standard values, which serve as reference points. 

So for instance lets's say I have a reaction flask and I put in 3.4 moles of A and 7.2 moles of be, and the reaction A ::equil:: B again: the ΔX° values (X = G, S, H) are the changes in X if I were to convert exactly 1 mol of A to exactly 1 mol of B at a given temperature (no equilibrium here - conversion is complete).  ΔX records what the changes in X are between the actual starting position in my real flask (3.4 moles A and 7.2 moles B) and the equilibrium point.  ΔX° values do not change if I change the starting conditions in my reaction flask (other than temperature, though see below) because they are reference values.  ΔX values WILL change if I change the starting conditions in my reaction flask.  For instance, if I start with 7.2 moles A and 3.4 moles B, where I am in relation to the equilibrium point will be very different, so the change in Gibbs energy, enthalpy and entropy are going to be different.   

The ΔX° values are derived from differences in the formation enthalpy/enthalpy/gibbs energy values for the reactants and products, individually, which are themselves derived from the change in X that occurs to form each reactant and product from the elemental precursors at standard state.  (That is, for our A to B reaction, ΔX° = ΔfX°(B) - ΔfX°(A).  If A is a hydrocarbon with formula CxHy, for example, ΔfX°(A) is in turn determined by measuring the change in X for the reaction C(s) + H2(g) :rarrow: CxHy or some such.  The enthalpy value, as an example, would usually determined by calorimetry experiments at a well controlled temperature - how much heat it takes to blast a known quantity of A back into its constituent pieces.  This is what I mean by standard values - they are determined under standard conditions and serve as reference points for real thermodynamical measurements.)  The standard Gibbs energy changes can't really be measured experimentally and are determined from the standard enthalpies and entropies, which CAN be measured experimentally.  Most of the time the standard enthalpy and entropy changes are measured at 298.15.  Therefore the standard Gibbs energies can also only be determined directly at 298.15 K.  However as we've remarked the standard enthalpies and entropies are usually assumed to be temperature independent, so the standard Gibbs energy change at any temperature can be estimated by ΔG° = ΔH° - TΔS°.  If the standard enthalpy and/or entropy changes are known to NOT be temperature independent, then your life is harder and some other method will have to be used.

So, take home message: ΔG° is a standard value for a given reaction performed to completion on a specific, predetermined set of conditions.  ΔG is the energy change that occurs if I put some of the reactants and products in a pot and wait until equilibrium is reached. 

This is my understanding of it, anyway.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

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Re: Gibbs' Free energy Olympiad Question
« Reply #23 on: March 30, 2013, 07:49:14 AM »
So ΔG° actually is temperature dependent (I'm guessing the ° symbol then only refers strictly to pressure?), but since ΔH° and ΔS° are (approximated to be) temperature independent, then whatever T we want ΔG° for, we can just take the 298.15 K values of ΔH° and ΔS° (under the assumption that temperature modification is negligible) and then plug that in together with the T to ΔG°=ΔH°-T·ΔS°.
Yes that's pretty much it.  Forgetting the Δ for a moment, the ° in a G° indicates that it is the amount of free energy change resulting from the formation of a substance from the standard state of the component elements.  ΔG° for a reaction would effectively be the change in free energy occuring if the reactants in their standard states were converted into products in their standard states.  Pretty much every tabulated value you're going to find is going to be measured at 298.15.

Quote
However, this would suggest that, if we take the approximation that ΔH and ΔS are not temperature dependent and we are working at 1 bar pressure, we could also have ΔG=ΔH°-T·ΔS° - which would suddenly mean ΔG=ΔG° and this is far too great a generalization to be true when all we've said is that ΔH and ΔS are not temperature dependent (and we're at 1 bar pressure). Particularly as it means K=1 in all such cases, which obviously does not have to be true. So this problem persists!
No, here you are confusing the standard enthalpy/entropy changes, which are benchmark values, and the actual enthalpy/entropy changes that occur for a reaction given specific concentrations of reactants and products.  (And therefore you are also confusing ΔG and ΔG°.)

Think of it this way: For a reaction, the standard thermodynamic values represent the change in the thermodynamic values that would result when taking 1 mol of the reactants and converting to one mol of the products at the specified temperature.  This is why they are standard - they are values determined under a specific set of conditions that serve as reference values, because most thermodynamic quantities only have relevance when measured respective to something else.  It is change we are interested in, not absolute quantities.  However most chemical reactions in reality do not relate converting 1 mol of this exactly into one mol of that.  First, there are always equilibria - so conversion isn't complete - and you're likely to have concentrations or amounts of reacting substances that differ from 1 mol.  The DG, DH and DS for a reaction are the actual changes in those values that occur based on the specified conditions.  Those valuese are reported relative to the standard values, which serve as reference points. 

So for instance lets's say I have a reaction flask and I put in 3.4 moles of A and 7.2 moles of be, and the reaction A ::equil:: B again: the ΔX° values (X = G, S, H) are the changes in X if I were to convert exactly 1 mol of A to exactly 1 mol of B at a given temperature (no equilibrium here - conversion is complete).  ΔX records what the changes in X are between the actual starting position in my real flask (3.4 moles A and 7.2 moles B) and the equilibrium point.  ΔX° values do not change if I change the starting conditions in my reaction flask (other than temperature, though see below) because they are reference values.  ΔX values WILL change if I change the starting conditions in my reaction flask.  For instance, if I start with 7.2 moles A and 3.4 moles B, where I am in relation to the equilibrium point will be very different, so the change in Gibbs energy, enthalpy and entropy are going to be different.   

The ΔX° values are derived from differences in the formation enthalpy/enthalpy/gibbs energy values for the reactants and products, individually, which are themselves derived from the change in X that occurs to form each reactant and product from the elemental precursors at standard state.  (That is, for our A to B reaction, ΔX° = ΔfX°(B) - ΔfX°(A).  If A is a hydrocarbon with formula CxHy, for example, ΔfX°(A) is in turn determined by measuring the change in X for the reaction C(s) + H2(g) :rarrow: CxHy or some such.  The enthalpy value, as an example, would usually determined by calorimetry experiments at a well controlled temperature - how much heat it takes to blast a known quantity of A back into its constituent pieces.  This is what I mean by standard values - they are determined under standard conditions and serve as reference points for real thermodynamical measurements.)  The standard Gibbs energy changes can't really be measured experimentally and are determined from the standard enthalpies and entropies, which CAN be measured experimentally.  Most of the time the standard enthalpy and entropy changes are measured at 298.15.  Therefore the standard Gibbs energies can also only be determined directly at 298.15 K.  However as we've remarked the standard enthalpies and entropies are usually assumed to be temperature independent, so the standard Gibbs energy change at any temperature can be estimated by ΔG° = ΔH° - TΔS°.  If the standard enthalpy and/or entropy changes are known to NOT be temperature independent, then your life is harder and some other method will have to be used.

So, take home message: ΔG° is a standard value for a given reaction performed to completion on a specific, predetermined set of conditions.  ΔG is the energy change that occurs if I put some of the reactants and products in a pot and wait until equilibrium is reached. 

This is my understanding of it, anyway.

I think I understand now! Thank you! :)

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