From owner-chemistry@ccl.net Wed Feb 15 08:33:00 2017 From: "Ernest Chamot echamot-*-chamotlabs.com" To: CCL Subject: CCL: Entropy of a Bimolecular System Message-Id: <-52640-170214154113-22022-OwiOExNb+nAwucAuye5EUw_+_server.ccl.net> X-Original-From: Ernest Chamot Content-Type: multipart/alternative; boundary="Apple-Mail=_DF778CD4-98D0-42A1-A31E-55749D914DC6" Date: Tue, 14 Feb 2017 14:41:04 -0600 Mime-Version: 1.0 (Mac OS X Mail 10.2 \(3259\)) Sent to CCL by: Ernest Chamot [echamot__chamotlabs.com] --Apple-Mail=_DF778CD4-98D0-42A1-A31E-55749D914DC6 Content-Transfer-Encoding: quoted-printable Content-Type: text/plain; charset=utf-8 Hi Per-Ola, First, thanks to all for the assistance from a lot of different = perspectives. It helps refresh my thinking about this. Several replies noted that I do, indeed, need to be thinking about the 3 = rotational and 3 translational degrees of freedom of the two separate = molecules, and reiterated how those terms are typically handled: either = ignored (bad) or approximated, by particle-in-a-box for translation, and = rigid rotor for rotation. Your reply, and a discussion with a colleague, clarified things for me: > It all hangs on the degrees of freedom, which are the same for a given = number of atoms, irrespective of whether it=E2=80=99s two infinitely = separated molecules or a tight complex (unless it becomes linear). What = happens when they meet is that six degrees of freedom are changed from = rotation+translation to vibration. The QM packages I use include the = rot+trans degrees of freedom in the final free energy, so that the final = numbers yield the same number of degrees of freedom, and can be added. My confusion had boiled down to thinking of: A -> B + C with, =E2=88=86G(rxn) =3D G(B) + G(C) - G(A) where each calculated free energy, either included rotational and = translational corrections for the entropy, or only calculated the = vibrational entropy within the frame of reference of the molecule. I do individual molecule calculations that incorporate entropy = corrections for the 3 translational and 3 rotational degrees of freedom = in calculating =E2=88=86G(rxn), so I am adding 6 corrections in = calculating G(A), but 12 corrections in calculating the products: 6 in = G(B) and 6 in G(C). So it seemed like I would either be double counting = if I used corrections, or be ignoring 6 more degrees of freedom in the = products than in the reactants, if I didn=E2=80=99t include corrections. As I understand it now, though, the entropy difference from fewer = rotational and translational correction terms in the reactant, is = balanced by the additional vibrational entropy in the reactant, with = more vibrational modes than in the product molecules. So it is correct = to use the translational and rotational entropy corrections, and = subtract the corrected reactant free energy, G(A), from the sum of the = corrected product free energies, G(B) and G(C), to get the =E2=88=86G of = reaction. =20 Thanks all. EC Ernest Chamot Chamot Labs, Inc. http://www.chamotlabs.com --Apple-Mail=_DF778CD4-98D0-42A1-A31E-55749D914DC6 Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset=utf-8
Hi Per-Ola,

First, thanks to all for the assistance from a lot of = different perspectives.  It helps refresh my thinking about = this.

Several = replies noted that I do, indeed, need to be thinking about the 3 = rotational and 3 translational degrees of freedom of the two separate = molecules, and reiterated how those terms are typically handled: either = ignored (bad) or approximated, by particle-in-a-box for translation, and = rigid rotor for rotation.

Your reply, and a discussion with a colleague, clarified = things for me:

It all hangs = on the degrees of freedom, which are the same for a given number of = atoms, irrespective of whether it=E2=80=99s two infinitely separated = molecules or a tight complex (unless it becomes linear). What happens = when they meet is that six degrees of freedom are changed from = rotation+translation to vibration. The QM packages I use include the = rot+trans degrees of freedom in the final free energy, so that the final = numbers yield the same number of degrees of freedom, and can be = added.

My confusion had boiled = down to thinking of:

  A -> B + C

with,

 =E2=88=86G(rxn) =3D G(B) + G(C) - = G(A)

where = each calculated free energy, either included rotational and = translational corrections for the entropy, or only calculated the = vibrational entropy within the frame of reference of the = molecule.

I do = individual molecule calculations that incorporate entropy corrections = for the 3 translational and 3 rotational degrees of freedom in = calculating =E2=88=86G(rxn), so I am adding 6 corrections in calculating = G(A), but 12 corrections in calculating the products: 6 in G(B) and 6 in = G(C).  So it seemed like I would either be double counting if I = used corrections, or be ignoring 6 more degrees of freedom in the = products than in the reactants, if I didn=E2=80=99t include = corrections.

As = I understand it now, though, the entropy difference from fewer = rotational and translational correction terms in the reactant, is = balanced by the additional vibrational entropy in the reactant, with = more vibrational modes than in the product molecules.  So it is = correct to use the translational and rotational entropy corrections, and = subtract the corrected reactant free energy, G(A), from the sum of the = corrected product free energies, G(B) and G(C), to get the =E2=88=86G of = reaction.  

Thanks all.

EC

Ernest Chamot
Chamot Labs, Inc.
http://www.chamotlabs.com


= --Apple-Mail=_DF778CD4-98D0-42A1-A31E-55749D914DC6-- From owner-chemistry@ccl.net Wed Feb 15 10:48:01 2017 From: "Michael Gilson mgilson|-|ucsd.edu" To: CCL Subject: CCL: Entropy of a Bimolecular System Message-Id: <-52641-170215100904-9996-IO6syXcCW4gL/FjpLwEPvg||server.ccl.net> X-Original-From: Michael Gilson Content-Type: multipart/alternative; boundary="------------273E0B6EC55B394245D17688" Date: Wed, 15 Feb 2017 07:08:34 -0800 MIME-Version: 1.0 Sent to CCL by: Michael Gilson [mgilson*ucsd.edu] This is a multi-part message in MIME format. --------------273E0B6EC55B394245D17688 Content-Type: text/plain; charset=utf-8; format=flowed Content-Transfer-Encoding: 8bit Hi Ernest, Exactly. Your point about rot/trans degrees of freedom being converted to vib degrees of freedom in the context goes back to a classic paper > from Scheraga, using the classical approximation of stat thermo: http://www.jbc.org/content/238/1/172.full.pdf Regards, Mike On 2/14/2017 12:41 PM, Ernest Chamot echamot-*-chamotlabs.com wrote: > Hi Per-Ola, > > First, thanks to all for the assistance from a lot of different > perspectives. It helps refresh my thinking about this. > > Several replies noted that I do, indeed, need to be thinking about the > 3 rotational and 3 translational degrees of freedom of the two > separate molecules, and reiterated how those terms are typically > handled: either ignored (bad) or approximated, by particle-in-a-box > for translation, and rigid rotor for rotation. > > Your reply, and a discussion with a colleague, clarified things for me: > >> It all hangs on the degrees of freedom, which are the same for a >> given number of atoms, irrespective of whether it’s two infinitely >> separated molecules or a tight complex (unless it becomes linear). >> What happens when they meet is that six degrees of freedom are >> changed from rotation+translation to vibration. The QM packages I use >> include the rot+trans degrees of freedom in the final free energy, so >> that the final numbers yield the same number of degrees of freedom, >> and can be added. > > My confusion had boiled down to thinking of: > > A -> B + C > > with, > > ∆G(rxn) = G(B) + G(C) - G(A) > > where each calculated free energy, either included rotational and > translational corrections for the entropy, or only calculated the > vibrational entropy within the frame of reference of the molecule. > > I do individual molecule calculations that incorporate entropy > corrections for the 3 translational and 3 rotational degrees of > freedom in calculating ∆G(rxn), so I am adding 6 corrections in > calculating G(A), but 12 corrections in calculating the products: 6 in > G(B) and 6 in G(C). So it seemed like I would either be double > counting if I used corrections, or be ignoring 6 more degrees of > freedom in the products than in the reactants, if I didn’t include > corrections. > > As I understand it now, though, the entropy difference from fewer > rotational and translational correction terms in the reactant, is > balanced by the additional vibrational entropy in the reactant, with > more vibrational modes than in the product molecules. So it is > correct to use the translational and rotational entropy corrections, > and subtract the corrected reactant free energy, G(A), from the sum of > the corrected product free energies, G(B) and G(C), to get the ∆G of > reaction. > > Thanks all. > > EC > > Ernest Chamot > Chamot Labs, Inc. > http://www.chamotlabs.com > > --------------273E0B6EC55B394245D17688 Content-Type: text/html; charset=utf-8 Content-Transfer-Encoding: 8bit Hi Ernest,

Exactly. Your point about rot/trans degrees of freedom being converted to vib degrees of freedom in the context goes back to a classic paper from Scheraga, using the classical approximation of stat thermo: http://www.jbc.org/content/238/1/172.full.pdf

Regards,
Mike

On 2/14/2017 12:41 PM, Ernest Chamot echamot-*-chamotlabs.com wrote:
Hi Per-Ola,

First, thanks to all for the assistance from a lot of different perspectives.  It helps refresh my thinking about this.

Several replies noted that I do, indeed, need to be thinking about the 3 rotational and 3 translational degrees of freedom of the two separate molecules, and reiterated how those terms are typically handled: either ignored (bad) or approximated, by particle-in-a-box for translation, and rigid rotor for rotation.

Your reply, and a discussion with a colleague, clarified things for me:

It all hangs on the degrees of freedom, which are the same for a given number of atoms, irrespective of whether it’s two infinitely separated molecules or a tight complex (unless it becomes linear). What happens when they meet is that six degrees of freedom are changed from rotation+translation to vibration. The QM packages I use include the rot+trans degrees of freedom in the final free energy, so that the final numbers yield the same number of degrees of freedom, and can be added.

My confusion had boiled down to thinking of:

  A -> B + C

with,

 ∆G(rxn) = G(B) + G(C) - G(A)

where each calculated free energy, either included rotational and translational corrections for the entropy, or only calculated the vibrational entropy within the frame of reference of the molecule.

I do individual molecule calculations that incorporate entropy corrections for the 3 translational and 3 rotational degrees of freedom in calculating ∆G(rxn), so I am adding 6 corrections in calculating G(A), but 12 corrections in calculating the products: 6 in G(B) and 6 in G(C).  So it seemed like I would either be double counting if I used corrections, or be ignoring 6 more degrees of freedom in the products than in the reactants, if I didn’t include corrections.

As I understand it now, though, the entropy difference from fewer rotational and translational correction terms in the reactant, is balanced by the additional vibrational entropy in the reactant, with more vibrational modes than in the product molecules.  So it is correct to use the translational and rotational entropy corrections, and subtract the corrected reactant free energy, G(A), from the sum of the corrected product free energies, G(B) and G(C), to get the ∆G of reaction.  

Thanks all.

EC

Ernest Chamot
Chamot Labs, Inc.



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