July 07, 2020, 11:33:06 AM
Forum Rules: Read This Before Posting


Topic: Gibbs' Free energy Olympiad Question  (Read 12936 times)

0 Members and 1 Guest are viewing this topic.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Gibbs' Free energy Olympiad Question
« 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?

Offline XGen

  • Full Member
  • ****
  • Posts: 127
  • Mole Snacks: +9/-4
Re: Gibbs' Free energy Olympiad Question
« Reply #1 on: March 17, 2013, 12:39:42 PM »
It probably assumes enthalpy of reaction and entropy have negligible change at different temperatures.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' Free energy Olympiad Question
« Reply #2 on: March 17, 2013, 01:01:13 PM »
It probably assumes enthalpy of reaction and entropy have negligible change at different temperatures.

Elaborate please.

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).

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' Free energy Olympiad Question
« Reply #3 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).

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 2895
  • Mole Snacks: +451/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' Free energy Olympiad Question
« Reply #4 on: March 28, 2013, 11:45:41 AM »
Typo perhaps?
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

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' Free energy Olympiad Question
« Reply #5 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.
« Last Edit: March 28, 2013, 12:23:01 PM by Big-Daddy »

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 2895
  • Mole Snacks: +451/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' Free energy Olympiad Question
« Reply #6 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.
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 Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 2895
  • Mole Snacks: +451/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' Free energy Olympiad Question
« Reply #7 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.
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

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' Free energy Olympiad Question
« Reply #8 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

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 2895
  • Mole Snacks: +451/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' Free energy Olympiad Question
« Reply #9 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. 
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

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' Free energy Olympiad Question
« Reply #10 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.).

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 2895
  • Mole Snacks: +451/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' Free energy Olympiad Question
« Reply #11 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.
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

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' Free energy Olympiad Question
« Reply #12 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?

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 2895
  • Mole Snacks: +451/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' Free energy Olympiad Question
« Reply #13 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.
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

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' Free energy Olympiad Question
« Reply #14 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?

Sponsored Links