April 19, 2024, 02:31:14 AM
Forum Rules: Read This Before Posting


Topic: Enthalpy change = 0  (Read 15833 times)

0 Members and 1 Guest are viewing this topic.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Enthalpy change = 0
« Reply #15 on: April 29, 2013, 01:35:49 PM »
I just read the IChO problem. Nothing wrong with that problem at all. But there's critical assumptions there that they explicitly mention that you left out.

I quote verbatim:

(1) The difference in the chemical behaviour of various isotopes is usually negligible, unless the relative change in the molecular mass is considerable.

(2) What are the mole fractions of these species in bromine at natural abundance

(3) What is the standard molar entropy change associated with this reaction, supposing that the chemical behaviour of the molecules involved is identical?

With those assumptions everything's fine! But those are assumptions. Givens. Not something that can be inferred.

All of @corribus's great exposition on ZPE's etc. was exactly what Point #1 was trying to tell you. So also the bit about natural abundance.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Enthalpy change = 0
« Reply #16 on: April 29, 2013, 02:09:08 PM »
I just read the IChO problem. Nothing wrong with that problem at all. But there's critical assumptions there that they explicitly mention that you left out.

I quote verbatim:

(1) The difference in the chemical behaviour of various isotopes is usually negligible, unless the relative change in the molecular mass is considerable.

(2) What are the mole fractions of these species in bromine at natural abundance

(3) What is the standard molar entropy change associated with this reaction, supposing that the chemical behaviour of the molecules involved is identical?

With those assumptions everything's fine! But those are assumptions. Givens. Not something that can be inferred.

All of @corribus's great exposition on ZPE's etc. was exactly what Point #1 was trying to tell you. So also the bit about natural abundance.

I see, sorry. I should have dropped in the problem verbatim (reason I avoided this is because I only needed help with a later part, and didn't want you to be discouraged from helping).

By the same logic, though, my point about K for the reaction I listed in the OP being 4 also works, at natural abundance. Indeed, for any isotopic exchange reaction, it is 4 at natural abundance. Problems with the logic in there? I suppose that was something which I failed to state, but I was aware of.

1) and 3) raises my issue. "the chemical behaviour of the molecules involved is identical" can you explain how this necessarily means that ΔHr°=0? Because it's that which I don't understand.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Enthalpy change = 0
« Reply #17 on: April 29, 2013, 03:37:16 PM »

1) and 3) raises my issue. "the chemical behaviour of the molecules involved is identical" can you explain how this necessarily means that ΔHr°=0? Because it's that which I don't understand.

Your reaction's of the form:

A2 + B2  ::equil:: 2 AB

If A=B (=X , say) (follows from " chemical behaviour of the molecules involved is identical" ) then the reaction becomes:

X2 + X2   ::equil:: 2 XX

i.e.

2 X2  ::equil:: 2 X2

This, almost tautologically, has to have ΔHr°=0. (or for that matter ΔY =0 where Y is essentially any "strictly chemical" property. I'm not so sure about the last one though. Let's see if someone objects.. )

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Enthalpy change = 0
« Reply #18 on: April 29, 2013, 04:39:22 PM »

1) and 3) raises my issue. "the chemical behaviour of the molecules involved is identical" can you explain how this necessarily means that ΔHr°=0? Because it's that which I don't understand.

Your reaction's of the form:

A2 + B2  ::equil:: 2 AB

If A=B (=X , say) (follows from " chemical behaviour of the molecules involved is identical" ) then the reaction becomes:

X2 + X2   ::equil:: 2 XX

i.e.

2 X2  ::equil:: 2 X2

This, almost tautologically, has to have ΔHr°=0. (or for that matter ΔY =0 where Y is essentially any "strictly chemical" property. I'm not so sure about the last one though. Let's see if someone objects.. )

Ah ok ... so if the chemical behaviour is said to be equal then all the molecules can be called the same thing. In that case of course there is no change happening. However if chemical behaviour being equal means we can write down all the molecules as the same to reach 2 X2  ::equil:: 2 X2, then our entropy change should also be 0. I thought the physical properties might not be included when it says "chemical behaviour is equal" - so then do the physical properties have no relationship (negligible) to enthalpy, meaning that only chemical properties affect enthalpy (to within a good approximation)?

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3481
  • Mole Snacks: +530/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Enthalpy change = 0
« Reply #19 on: April 29, 2013, 06:06:18 PM »
And the equilibrium constant MUST be 1 in that case.  (Because ΔG0 = 0)
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: Enthalpy change = 0
« Reply #20 on: April 29, 2013, 06:25:07 PM »
And the equilibrium constant MUST be 1 in that case.  (Because ΔG0 = 0)

In the case where chemical properties are the same across all molecules so ΔH°=0, and ΔS°=0 (i.e. physical properties are the same), you mean? Whereas the calculation will go as in the IChO problem solution if only ΔH°=0 as only chemical properties are the same (differences are negligible).

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3481
  • Mole Snacks: +530/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Enthalpy change = 0
« Reply #21 on: April 29, 2013, 07:59:24 PM »
As I said in my earlier, longer post while ΔH isn't exactly zero, it's not a terrible approximation.  However it IS a terrible approximation for the entropic term.  But I don't think I'd make a distinction that one represents chemical properties and one physical properties.  At this point, what is the difference between chemical and physical? 
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 curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Enthalpy change = 0
« Reply #22 on: April 30, 2013, 01:29:09 AM »
As I said in my earlier, longer post while ΔH isn't exactly zero, it's not a terrible approximation.  However it IS a terrible approximation for the entropic term.  But I don't think I'd make a distinction that one represents chemical properties and one physical properties.  At this point, what is the difference between chemical and physical?

I think there is a distinction and that's what makes ΔH=0 and ΔS≠0.

Let's assume perfect chemical identity. That's not a bad assumption for something like Cl or Br with a large electronic shell. Let's also assume ZPE changes are negligible. That's mean no differences in any sort of bonding, attraction etc. Ergo ΔH=0.

But K=1 is not a good assumption. (I think) In fact, look at experimental values of K for diatomic isotopic exchange. Except H where ZPE etc. deviations are relatively large, K is indeed always very close to 4. Not 1. Empirically.

Hence I think it's perfectly reasonable to use ΔH=0 and ΔS≠0 both at the same time. Hence ΔG≠0.

The entropic question is an interesting one. I think that is a variant of the Gibbs paradox. Normally Cl2 molecules are indistinguishable. That is key to the derivation that tells us that mixing two containers of Cl2 will lead to no entropy change (unlike mixing two different gases where there indeed is an entropy of mixing).

Opening or closing a shutter between two partitions of Cl2 leaves the system unchanged. (since there is no way to label Chlorine molecules)

OTOH isotopic labeling allows us a way to make Cl2 molecules distinguishable! That minor change by itself is sufficient to lead to an entropy of mixing. Distinguishibility matters. In a way it's a bit weird and counter intuitive. It's like just being able to color some molecules differently will change entropy even if mass, forces and all other interactions remain the same.

Again, I'm a tad speculative here. I might be wrong.

PS. On second thoughts the physical / chemical distinction I made earlier is arbitrary. I just found the fact that Entropy can change merely be labeling very interesting.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3481
  • Mole Snacks: +530/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Enthalpy change = 0
« Reply #23 on: April 30, 2013, 08:33:40 AM »
What I meant was - calling enthalpy "chemical property" and entropy "physical property" is a strange distinction to make, because entropy obviously affects the outcomes of chemical change (i.e., reaction).  I agree with pretty much everything else you wrote.
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 curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Enthalpy change = 0
« Reply #24 on: April 30, 2013, 08:34:59 AM »
What I meant was - calling enthalpy "chemical property" and entropy "physical property" is a strange distinction to make, because entropy obviously affects the outcomes of chemical change (i.e., reaction). 

Agree.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Enthalpy change = 0
« Reply #25 on: April 30, 2013, 08:39:00 AM »
As an aside, it'd be interesting to try and get the same answer for ΔS using a more direct approach. i.e. counting possible configurations etc.

The equilibrium approach is an interesting shortcut.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Enthalpy change = 0
« Reply #26 on: April 30, 2013, 12:13:25 PM »
What I meant was - calling enthalpy "chemical property" and entropy "physical property" is a strange distinction to make, because entropy obviously affects the outcomes of chemical change (i.e., reaction).  I agree with pretty much everything else you wrote.

How about enthalpy being the "chemical properties" and entropy "both physical and chemical properties"? In other words, only if there really is no reaction going on is the entropy change 0 (i.e. A  ::equil:: A), whereas the enthalpy change is 0 to within a good approximation if the constituents are merely chemically equivalent?

I'm not sure what else to take away from this problem ...

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Enthalpy change = 0
« Reply #27 on: April 30, 2013, 02:36:45 PM »
Also, if we were looking at the molar fraction of ozone at natural abundance. Let's say we specify that 85% of O is 16O and 15% of O is 18O isotopes. Then is it reasonable to say that the abundance of [18O-16O-18O], the molecule with 2 18O atoms and 1 16O atom, will be (3!/2!)*0.152*0.85, and that of [16O-18O-18O] will be (3!/2!)*0.852*0.15? It seems to be a stats problem of how many arrangements there are of each molecule, multiplied by each atom in the molecule which has been raised to the power of how many times it shows up in the molecule (if we're looking at the bare maths).

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Enthalpy change = 0
« Reply #28 on: April 30, 2013, 03:06:10 PM »
Also, if we were looking at the molar fraction of ozone at natural abundance. Let's say we specify that 85% of O is 16O and 15% of O is 18O isotopes. Then is it reasonable to say that the abundance of [18O-16O-18O], the molecule with 2 18O atoms and 1 16O atom, will be (3!/2!)*0.152*0.85, and that of [16O-18O-18O] will be (3!/2!)*0.852*0.15? It seems to be a stats problem of how many arrangements there are of each molecule, multiplied by each atom in the molecule which has been raised to the power of how many times it shows up in the molecule (if we're looking at the bare maths).

Yes, I think.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Enthalpy change = 0
« Reply #29 on: April 30, 2013, 03:07:56 PM »
How about enthalpy being the "chemical properties" and entropy "both physical and chemical properties"? In other words, only if there really is no reaction going on is the entropy change 0 (i.e. A  ::equil:: A), whereas the enthalpy change is 0 to within a good approximation if the constituents are merely chemically equivalent?


Semantics. It's confusing, probably wrong and very unlikely to be useful.

Quote
I'm not sure what else to take away from this problem ...

I say you've pretty much flogged it for all it's worth.

Sponsored Links