# Chemical Forums

## Chemistry Forums for Students => High School Chemistry Forum => Chemistry Olympiad and other competitions => Topic started by: Big-Daddy on March 17, 2013, 11:11:06 AM

Title: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy on March 17, 2013, 11:11:06 AM
This problem is from the International Chemistry Olympiad 2012 preparatory set (Problem 5) and seems decently basic enough to be in the High School forum.

Given the thermodynamic data below, calculate the temperature at which gray Sn is in
equilibrium with white Sn (at 1 bar = 105 Pa pressure).

The ΔHfO and SO values of gray and white Sn are then given. Nothing else.

----------------------

The equilibrium is ostensibly Sn (s, gray)  ::equil::  Sn (s, white). The direction makes no difference.
But the solution lists the condition for equilibrium as ΔGrO=0. Then, of course, ΔHrO-T*ΔSrO=0 where ΔHrO=ΔHfO[Sn (white)]-ΔHfO[Sn (gray)] and ΔSrO=SO[Sn (white)]-SO[Sn (gray)]. The only trick to remember when solving is to convert ΔH values from kJ·mol-1 to J·mol-1. We can get T easily.

BUT: I thought the condition for equilibrium to be reached is ΔGr=0, not ΔGrO=0, which is an important distinction as ΔGrO+RTloge(Kc)=0 and so ΔGrO=0 does not leave a flexible result (it forces Kc=1), how is this right?
Title: Re: Gibbs' Free energy Olympiad Question
Post by: XGen on March 17, 2013, 12:39:42 PM
It probably assumes enthalpy of reaction and entropy have negligible change at different temperatures.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy on March 17, 2013, 01:01:13 PM
It probably assumes enthalpy of reaction and entropy have negligible change at different temperatures.

Three equations here:
ΔGrO=ΔHrO-T*ΔSrO
ΔGr=ΔHr-T*ΔSr (The only difference is that this second case refers to the specific temperature and pressure at which you are operating, whereas the above case refers to the standard temperature and pressure, 298 K and 1 bar. We can find the standard case from standard enthalpies of formation and entropies, but not the second, condition-variant case.)

ΔGr=ΔGrO+R·T·loge(Q)

So what is the condition when the reaction reaches equilibrium? ΔGr=0, I hope. To take the approximation that ΔGrO=ΔGr and so ΔGrO=0 at equilibrium should then be tantamount to saying either T or loge(Q) is 0, so Q=1 (which is not a constraint we can place on the system) or T = 0 K (obviously not true).
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy on March 28, 2013, 11:10:17 AM
Can anyone help? The central question - why the solution seems to say ΔGrO=0 at equilibrium rather than ΔGr=0 - is still unanswered, because if they equal each other then the equation ΔGr=ΔGrO+R*T*loge(K) would force a value of 1 on K which we cannot allow unless we are told "there is the same number of molecules of Sn (white) as of Sn (gray) when the system reaches equilibrium" (we are not told this).
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Corribus on March 28, 2013, 11:45:41 AM
Typo perhaps?
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy on March 28, 2013, 12:04:16 PM
Typo perhaps?

But then how can we solve it? We don't know the equilibrium constant ... if we did then ΔGrO=ΔHrO-T·ΔSrO and ΔGrO=-R·T·loge(Kc) at equilibrium so ΔHrO-T·ΔSrO=-R·T·loge(Kc) and we have something solvable (given that ΔHrO and ΔSrO can be expressed easily in terms of enthalpies of formation and standard entropes) into T=ΔHrO/(ΔSrO-R·loge(Kc)). But without the Kc in that expression, I don't know how to do it ...?

The whole of the solution's method hinges around ΔGrO=0 and thereby being able to say ΔHrO-T·ΔSrO=0; now just avoid the usual trap of not putting ΔHrO in J·mol-1 and the answer comes out as T=ΔHrO/ΔSrO. Once I knew ΔGrO=0, only then could I solve it - but this doesn't seem at all obvious.

Maybe there is something implicit in the rest of the question to make this the case? Here's the full text (I should warn the IChO problems, particularly prep ones like this, usually start with long and often irrelevant preamble):

The ductility and malleability typical of metals has made metals essential structural elements in modern construction. The thermodynamically stable form of elemental tin at 298 K and ambient pressure is white tin, which has mechanical properties typical of metals and therefore can be used as a building material. At lower temperatures, however, a second allotrope of tin, gray tin, becomes thermodynamically stable. Because gray tin is much more brittle than white tin, structural elements made of tin that are kept at low temperatures for prolonged periods may crumble and fail. Because this failure resembles a disease, it has been termed the "tin pest".

a) Given the thermodynamic data below, calculate the temperature at which gray Sn is in
equilibrium with white Sn (at 1 bar = 105 Pa pressure).

Followed by the standard enthalpies of formation and standard entropies. Writing out the equilibrium Sn (s, gray)  ::equil:: Sn (s, white) is not actually done for you but is obvious.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Corribus on March 28, 2013, 12:07:08 PM
It's hard to offer much guidance without the complete wording of the problem.  Summarizing what is asked for can lead to things being missed.  Why don't you write out the entire question, with all data provided, then I will take a look.

EDIT: Never mind, I see you did that in the reponse.  I only read the first sentence. :P  Give me some time, I will look.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Corribus on March 28, 2013, 12:46:06 PM
Can you provide the thermodynamic data they provided?  I can look it up, but I'd prefer to work with the same values.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy on March 28, 2013, 01:12:13 PM
Can you provide the thermodynamic data they provided?  I can look it up, but I'd prefer to work with the same values.

Substance: ΔHfO /(kJ·mol-1); SO /(J·mol-1·K-1)
Sn (s, gray): 2.016; 44.14
Sn (s, white): 0.000; 51.18
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Corribus on March 28, 2013, 03:19:55 PM
Yeah, so this one threw me for a little, because the wording is really vague.  What's throwing me is their use of the word "equilibrium".  Given time, there is an equilibrium reached at ANY temperature.  The interconversion between the two forms of tin is well known, and while the metallic form is prevalent at higher temperatures, the equilibrium tends to favor gray tin when it gets cold.  (Gray tin is brittle and nonmetallic, and it has been theorized that the reason Napoleon lost to the Russians was because the buttons holding his soldiers' uniforms together were made of tin and didn't survive the cold Russian weather; that seems a bit simple to me as an explanation, but it's a fun idea that chemistry beat the Grande Armee...)  There is a point, then, where the the equilibrium favors neither form, that is - where the concentration of each is the same at equilibrium.  I have a feeling that this is what the problem means by equilibrium - that is, the "temperature at which gray Sn is in equilibrium with white Sn" is not referring to chemical equilibrium (where the rate forward equals the rate backward) but where the two concentrations are the same when chemical equilibrium is reached.  Otherwise it really makes no sense, because chemical equilibrium can be reached for any temperature - the temperature just impacts the relative concentrations of the two forms when equilibrium is reached.

I hope that makes sense.  And if my suspicion is true, that's an awfully poor choice of wording to use.  I don't know who wrote the problem but if it's intended for international audiences, it's very possible it was originally written in another language and then translated, possibly by someone with little knowledge in chemistry.

Anyway, never mind all that.

What the question I think is asking is for you to find the temperature at which gray Sn becomes the dominant form, that is, at chemical equilibrium, when [Sn,white] = [Sn,gray].  Needless to say, in this instance the equilibrium constant K is going to equal 1.

So: At what temperature does K = 1?

I think you've figured it out from here.  That's an easy justification for why ΔG° according to the problem solution is also 0.  Notice if you calculate ΔG° at 298.15 from your values, you do get an answer very close to 0 (-0.007 kJ/mol), but be careful here because every source I have lists the ΔH0 for gray Tin as around -2.1, not +2.1, AND when you calculate ΔG° this way you need to use 298.15 K, not the mystery temperature you're trying to solve).

I think that will give you the right answer (using your numbers I got about 286 K, which is close to the experimental value; interesting when I use values from the CRC, I got a value that's way off.  Another indication this question is bad.), but I've got more issues with this problem beyond those I've already described. I hate problems that rely on you making assumptions that you maybe wouldn't otherwise expect to make, but this is what this problem is requiring.  You are also implicitly assuming that ΔH and ΔS are independent of temperature.  So the ΔH and ΔS values you calculate from the heats of formation are appropriate at all temperatures.  It's generally not a bad approximation, but good questions make it clear when you are supposed to make certain approximations.  Add that on top of the poor wording and you get a problem that's really confusing IMO.  I'm still confusing myself writing about it so I think I'll quit now.

In the end your logic was sound - it was the question that was screwy.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy on March 28, 2013, 03:40:46 PM
Yeah, so this one threw me for a little, because the wording is really vague.  What's throwing me is their use of the word "equilibrium".  Given time, there is an equilibrium reached at ANY temperature.  The interconversion between the two forms of tin is well known, and while the metallic form is prevalent at higher temperatures, the equilibrium tends to favor gray tin when it gets cold.  (Gray tin is brittle and nonmetallic, and it has been theorized that the reason Napoleon lost to the Russians was because the buttons holding his soldiers' uniforms together were made of tin and didn't survive the cold Russian weather; that seems a bit simple to me as an explanation, but it's a fun idea that chemistry beat the Grande Armee...)  There is a point, then, where the the equilibrium favors neither form, that is - where the concentration of each is the same at equilibrium.  I have a feeling that this is what the problem means by equilibrium - that is, the "temperature at which gray Sn is in equilibrium with white Sn" is not referring to chemical equilibrium (where the rate forward equals the rate backward) but where the two concentrations are the same when chemical equilibrium is reached.  Otherwise it really makes no sense, because chemical equilibrium can be reached for any temperature - the temperature just impacts the relative concentrations of the two forms when equilibrium is reached.

I hope that makes sense.  And if my suspicion is true, that's an awfully poor choice of wording to use.  I don't know who wrote the problem but if it's intended for international audiences, it's very possible it was originally written in another language and then translated, possibly by someone with little knowledge in chemistry.

Anyway, never mind all that.

What the question I think is asking is for you to find the temperature at which gray Sn becomes the dominant form, that is, at chemical equilibrium, when [Sn,white] = [Sn,gray].  Needless to say, in this instance the equilibrium constant K is going to equal 1.

So: At what temperature does K = 1?

I think you've figured it out from here.  That's an easy justification for why ΔG° according to the problem solution is also 0.  Notice if you calculate ΔG° at 298.15 from your values, you do get an answer very close to 0 (-0.007), but be careful here because every source I have lists the ΔH0 for gray Tin as around -2.1, not +2.1, AND when you calculate ΔG° this way you need to use 298.15 K, not the mystery temperature you're trying to solve).

I think that will give you the right answer (using your numbers I got about 286 K, which is close to the experimental value; interesting when I use values from the CRC, I got a value that's way off.  Another indication this question is bad.), but I've got more issues with this problem beyond those I've already described. I hate problems that rely on you making assumptions that you maybe wouldn't otherwise expect to make, but this is what this problem is requiring.  You are also implicitly assuming that ΔH and ΔS are independent of temperature.  So the ΔH and ΔS values you calculate from the heats of formation are appropriate at all temperatures.  It's generally not a bad approximation, but good questions make it clear when you are supposed to make certain approximations.  Add that on top of the poor wording and you get a problem that's really confusing IMO.  I'm still confusing myself writing about it so I think I'll quit now.

In the end your logic was sound - it was the question that was screwy.

Ah I see. Yes, if it is the case that the two concentrations were said to be equal to each other, then the calculation is fine (in fact I had done it already - substitute Kc=1 to show an equal ratio of products to reactants into T=ΔHrO/(ΔSrO-R·loge(Kc)), the formula I found in my last post).

Can you remind me of the definition of ΔGrO as opposed to ΔGr please? I know ΔGrO is the "standard Gibbs' free energy change" of the reaction, but it also equals ΔHrO-T·ΔSrO which means it is temperature dependent, and I thought standard conditions meant temperature independence (I thought standard conditions referred specifically to 298.15 K, 1 bar pressure, etc.).
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Corribus on March 28, 2013, 10:02:13 PM
ΔG° is temperature dependent in that it depends on what temperature ΔS° and ΔH° were determined at.  Usually this is 298.15 and 1 atm.  So if you put ΔS° and ΔH° in the usual equation and 298.15 for T, you should get ΔG°.  Make sense?  Although, I don't think this is usually the way it is experimentally determined.  But I could be wrong.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy on March 29, 2013, 07:32:26 AM
ΔG° is temperature dependent in that it depends on what temperature ΔS° and ΔH° were determined at.  Usually this is 298.15 and 1 atm.  So if you put ΔS° and ΔH° in the usual equation and 298.15 for T, you should get ΔG°.  Make sense?  Although, I don't think this is usually the way it is experimentally determined.  But I could be wrong.

But we usually write the temperature dependent ΔG°=ΔH°-T·ΔS°. At equilibrium you have ΔG°=-R·T·loge(Kc). So does the first relationship strictly mean ΔG°=ΔH°-298.15·ΔS°, and the second ΔG°(298.15 K)=-R·T·loge(Kc)?

The problem is this seems to go against my previous calculations and intuition too. If I had used ΔH°-298.15·ΔS° instead as a definition for ΔG°, ΔH°-298.15·ΔS°=-R·T·loge(Kc) so T=(ΔH°-298.15·ΔS°)/(-R·loge(Kc)) and when Kc=1 there is no solution (logb(1)=0 so we're dividing by 0).

If ΔG° were not temperature dependent why would it be written ΔH°-T·ΔS° at all?
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Corribus on March 29, 2013, 10:46:57 AM
I think in the end the problem is that the original problem is not a very good one.  It's written in a very confusing way.  If you put 298.15 in your equation with the enthalpy and entropy of formation changes, you'll get a change in Gibbs energy of formation of close to zero.  I think in this case it's just coincidental that this happens.

I think the way you would usually solve a problem like this is to solve for ΔG° using ΔH° and ΔS° and the standard temperature (298.15).  Then you could use that ΔG° value to solve for the equilibrium constant at any other temperature using ΔG° = -RT ln K.  OR you could use that ΔG° to solve for a ΔG (at any temperature T) resultant from any pre-specified relative concentration of products/reactants (by using Q).  I believe you would find this to be pretty straightforward.

In this particular case we get a ΔG° that is coincidentally almost zero, which just makes it unnecessarily confusing.  That the problem solution tells you to assume this makes it doubly confusing.

What the problem is asking you to do is to make the deduction that at equilibrium,

ΔG° = ΔH° - TΔS°  AND that ΔG° = - RT ln K at equilibrium.

Setting these equal gives you

-RT ln K =  ΔH° - TΔS°

And doing some rearranging

ln K = - ΔH°/RT + ΔS°/R

The question then gives you a bit of confusing language intending you to find the temperature at which the concentrations of the two tins are equal at equilibrium.  So K = 1.

Which means: ΔH°/T = ΔS°

And you can solve for T.

The CATCH is that you must assume that ΔH° and ΔS° are temperature independent (that is, ΔH and ΔS under any temperature conditions are always equal to the ΔH° and ΔS° provided at the outset).

I don't know if this answered your question or not.  Taking one more stab at it:

Quote
If ΔG° were not temperature dependent why would it be written ΔH°-T·ΔS° at all?

Because this is the way ΔG° is defined.  It's the Gibbs change for complete transformation of reactants to products at a specific temperature.  It's basically a reference value, since thermodynamic changes are all expressed relative to something else.  The Gibbs energy in general is related to the enthalpy and entropy changes (and at a given temperature), so one way to determine ΔG° is to calculate it from ΔH° and ΔS°  Using any other T than 298.15 doesn't really make sense, because this is the temperature at which ΔH° and ΔS° are determined.  T isn't really a variable here.  It's a constant.  Confusing, I know, but there it is.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy on March 29, 2013, 10:58:17 AM
Because this is the way ΔG° is defined.  It's the Gibbs change for complete transformation of reactants to products at a specific temperature.  It's basically a reference value, since thermodynamic changes are all expressed relative to something else.  The Gibbs energy in general is related to the enthalpy and entropy changes (and at a given temperature), so one way to determine ΔG° is to calculate it from ΔH° and ΔS°  Using any other T than 298.15 doesn't really make sense, because this is the temperature at which ΔH° and ΔS° are determined.  T isn't really a variable here.  It's a constant.  Confusing, I know, but there it is.

That doesn't really answer my problem though. This isn't directly related to the OP of this thread but rather to a proper understanding of what ΔG° means.

ΔG°=ΔH°-T·ΔS° - if we assume ΔH° and ΔS° to be temperature independent. And because ΔG° is standard, ΔG°=ΔH°-298.15·ΔS°.

But now let's try finding T at which, for example, K=1 (as the equilibrium constant is of course temperature dependent but - apparently - ΔG° is temperature independent, so constant for different values of T in this calculation):

ΔH°-298.15·ΔS°=-R·T·loge(K)

So

T=(ΔH°-298.15·ΔS°)/(-R·loge(K))

And the whole thing breaks down because loge(1)=0. I'm pretty sure this is not the format of the solution. Which means that in ΔG°=ΔH°-T·ΔS° we cannot substitute T=298.15 K (or any other constant value). Which would suggest that ΔG° is temperature dependent?
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Yggdrasil 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.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Corribus 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.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy 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°!
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Corribus 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. :)

Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy 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.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Corribus 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. ;)
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy 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.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Corribus 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.
Title: Re: Gibbs' Free energy Olympiad Question
Post by: Big-Daddy 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! :)