From owner-chemistry@ccl.net Mon Mar 6 05:52:00 2006 From: "Roger Kevin Robinson r.robinson : imperial.ac.uk" To: CCL Subject: CCL:G: Gaussian Scan Message-Id: <-31119-060306055056-29949-aeTrdN+rTgqQuOSQkFX43w(!)server.ccl.net> X-Original-From: Roger Kevin Robinson Content-Transfer-Encoding: 7bit Content-Type: text/plain; charset=ISO-8859-1; format=flowed Date: Mon, 06 Mar 2006 10:55:57 +0000 MIME-Version: 1.0 Sent to CCL by: Roger Kevin Robinson [r.robinson]~[imperial.ac.uk] Hi, Im trying to learn how to use the gaussian scan feature. I decided to start with rotating ethane. This is my input file. %chk=/home2/rkr79/gaussian/scans/C2H6-scan.chk %mem=6MW %nproc=1 # scan rb3lyp/6-31g(d,p) geom=connectivity nosym Title Card Required 0 1 C H 1 B1 H 1 B2 2 A1 H 1 B3 3 A2 2 D1 C 1 B4 3 A3 4 D2 H 5 B5 1 A4 3 D3 H 5 B6 1 A5 3 D4 H 5 B7 1 A6 3 D5 B1 1.09518168 B2 1.09518168 B3 1.09518168 B4 1.53011970 B5 1.09518168 B6 1.09518168 B7 1.09518168 A1 107.49430338 A2 107.49430328 A3 111.38236537 A4 111.38236533 A5 111.38236537 A6 111.38236537 D1 115.45609965 D2 122.27195017 D3 -59.99999997 23 15.0 D4 60.00000006 D5 -180.00000000 1 2 1.0 3 1.0 4 1.0 5 1.0 2 3 4 5 6 1.0 7 1.0 8 1.0 6 7 8 As you can see im trying to rotate the torsion D3 by 15 degrees 23 times. However Gaussian always crashs this is the message in the log file. 8 H 1.766342 1.766342 0.000000 Small interatomic distances encountered: 7 6 Problem with the distance matrix. Error termination via Lnk1e in /opt/gaussian03/g03/l202.exe at Mon Mar 6 11:46:01 2006. Job cpu time: 0 days 0 hours 1 minutes 9.9 seconds. File lengths (MBytes): RWF= 24 Int= 0 D2E= 0 Chk= 18 Scr= 1 Now i suppose this is pretty self explainatory. But how do i fix it ? im using B3LYP/6-31(d,p) because a paper I found was also using it. I found this example online. http://amber.scripps.edu/Questions/mail/285.html as far as i can see my .com file is pretty similar. Does any one have anyother tips for scanning ? Thanks Roger From owner-chemistry@ccl.net Mon Mar 6 08:02:00 2006 From: "Tarmo Tamm tarmo[]chem.ut.ee" To: CCL Subject: CCL:G: Gaussian Scan Message-Id: <-31120-060306075920-824-q+WUQD2h2q2VMdhuJYhQXA+*+server.ccl.net> X-Original-From: Tarmo Tamm Content-Disposition: inline Content-Type: text/plain; charset=us-ascii Date: Mon, 6 Mar 2006 13:59:11 +0200 Mime-Version: 1.0 Sent to CCL by: Tarmo Tamm [tarmo[]chem.ut.ee] Hi Looks like you are just rotating one hydrogen instead of the CH3 group. You should define the remaining 2 hydrogens depending on the one you are rotating, in order to make them move together not towards each other. Good luck Tarmo Tamm University of Tartu On Mar 06, Roger Kevin Robinson r.robinson : imperial.ac.uk wrote: > Sent to CCL by: Roger Kevin Robinson [r.robinson]~[imperial.ac.uk] > Hi, > > Im trying to learn how to use the gaussian scan feature. I decided > to start with rotating ethane. > This is my input file. > From owner-chemistry@ccl.net Mon Mar 6 09:14:00 2006 From: "Roger Kevin Robinson r.robinson++imperial.ac.uk" To: CCL Subject: CCL:G: Gaussian Scan Message-Id: <-31121-060306090858-27312-kv3W9qeHB1aDOlGaKk9M+w~~server.ccl.net> X-Original-From: Roger Kevin Robinson Content-Transfer-Encoding: 7bit Content-Type: text/plain; charset=ISO-8859-1; format=flowed Date: Mon, 06 Mar 2006 14:13:49 +0000 MIME-Version: 1.0 Sent to CCL by: Roger Kevin Robinson [r.robinson : imperial.ac.uk] Hi i tried setting each torsion individualy but unsprisingly it did this 16 -60.000000 23 15.0000 17 60.000000 23 15.0000 18 -180.000000 23 15.0000 A total of 13824 points will be computed. which is nt what i want at all and the same problem is still there. Do I need to use this keyword opt=z-matrix instead of scan I have tried this and runs through but it seems to calculate more energy values than i was expecting and it doesnt put them in a neat little table like scan does. Any reason why it would do this ? Is there any way of telling a whole group to rotate ? Roger Tarmo Tamm tarmo[]chem.ut.ee wrote: >Sent to CCL by: Tarmo Tamm [tarmo[]chem.ut.ee] >Hi > >Looks like you are just rotating one hydrogen instead of the CH3 group. >You should define the remaining 2 hydrogens depending on the one you are >rotating, in order to make them move together not towards each other. > >Good luck > >Tarmo Tamm >University of Tartu > > > >On Mar 06, Roger Kevin Robinson r.robinson : imperial.ac.uk wrote: > > >>Sent to CCL by: Roger Kevin Robinson [r.robinson]~[imperial.ac.uk] >>Hi, >> >> Im trying to learn how to use the gaussian scan feature. I decided >>to start with rotating ethane. >>This is my input file.> > > > > From owner-chemistry@ccl.net Mon Mar 6 09:49:01 2006 From: "Marius Retegan marius.retegan,,yahoo.com" To: CCL Subject: CCL:G: Gaussian Scan Message-Id: <-31122-060306073854-30994-yWOzGPw9yPL7qoCuEGDOdg,server.ccl.net> X-Original-From: Marius Retegan Content-Transfer-Encoding: 8bit Content-Type: multipart/alternative; boundary="0-629923880-1141645127=:45467" Date: Mon, 6 Mar 2006 03:38:47 -0800 (PST) MIME-Version: 1.0 Sent to CCL by: Marius Retegan [marius.retegan^^yahoo.com] --0-629923880-1141645127=:45467 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: 8bit Dear Roger I think that your problem is a missing "s", which stands for scan in the input. D3 -59.99999997 23 15.0, should be D3 -59.99999997 s 23 15.0. A good place to learn about scans and geometry optimizations is the page of Prof. Dr. Hendrik Zipse. http://www.cup.uni-muenchen.de/oc/zipse/compchem/topics.html I hope this helps. Marius "Roger Kevin Robinson r.robinson : imperial.ac.uk" wrote: Sent to CCL by: Roger Kevin Robinson [r.robinson]~[imperial.ac.uk] Hi, Im trying to learn how to use the gaussian scan feature. I decided to start with rotating ethane. This is my input file. %chk=/home2/rkr79/gaussian/scans/C2H6-scan.chk %mem=6MW %nproc=1 # scan rb3lyp/6-31g(d,p) geom=connectivity nosym Title Card Required 0 1 C H 1 B1 H 1 B2 2 A1 H 1 B3 3 A2 2 D1 C 1 B4 3 A3 4 D2 H 5 B5 1 A4 3 D3 H 5 B6 1 A5 3 D4 H 5 B7 1 A6 3 D5 B1 1.09518168 B2 1.09518168 B3 1.09518168 B4 1.53011970 B5 1.09518168 B6 1.09518168 B7 1.09518168 A1 107.49430338 A2 107.49430328 A3 111.38236537 A4 111.38236533 A5 111.38236537 A6 111.38236537 D1 115.45609965 D2 122.27195017 D3 -59.99999997 23 15.0 D4 60.00000006 D5 -180.00000000 1 2 1.0 3 1.0 4 1.0 5 1.0 2 3 4 5 6 1.0 7 1.0 8 1.0 6 7 8 As you can see im trying to rotate the torsion D3 by 15 degrees 23 times. However Gaussian always crashs this is the message in the log file. 8 H 1.766342 1.766342 0.000000 Small interatomic distances encountered: 7 6 Problem with the distance matrix. Error termination via Lnk1e in /opt/gaussian03/g03/l202.exe at Mon Mar 6 11:46:01 2006. Job cpu time: 0 days 0 hours 1 minutes 9.9 seconds. File lengths (MBytes): RWF= 24 Int= 0 D2E= 0 Chk= 18 Scr= 1 Now i suppose this is pretty self explainatory. But how do i fix it ? im using B3LYP/6-31(d,p) because a paper I found was also using it. I found this example online. http://amber.scripps.edu/Questions/mail/285.html as far as i can see my .com file is pretty similar. Does any one have anyother tips for scanning ? Thanks Rogerhttp://www.ccl.net/cgi-bin/ccl/send_ccl_messagehttp://www.ccl.net/chemistry/sub_unsub.shtmlhttp://www.ccl.net/spammers.txt--0-629923880-1141645127=:45467 Content-Type: text/html; charset=iso-8859-1 Content-Transfer-Encoding: 8bit Dear Roger

I think that your problem is a missing "s", which stands for scan in the input.
D3 -59.99999997 23 15.0, should be
D3 -59.99999997 s 23 15.0.
A good place to learn about scans and geometry optimizations is the page of  Prof. Dr.  Hendrik Zipse.
http://www.cup.uni-muenchen.de/oc/zipse/compchem/topics.html
I hope this helps.
Marius
"Roger Kevin Robinson r.robinson : imperial.ac.uk" <owner-chemistry[]ccl.net> wrote:
Sent to CCL by: Roger Kevin Robinson [r.robinson]~[imperial.ac.uk]
Hi,

Im trying to learn how to use the gaussian scan feature. I decided
to start with rotating ethane.
This is my input file.

%chk=/home2/rkr79/gaussian/scans/C2H6-scan.chk
%mem=6MW
%nproc=1
# scan rb3lyp/6-31g(d,p) geom=connectivity nosym

Title Card Required

0 1
C
H 1 B1
H 1 B2 2 A1
H 1 B3 3 A2
2 D1
C 1 B4 3 A3
4 D2
H 5 B5 1 A4
3 D3
H 5 B6 1 A5
3 D4
H 5 B7 1 A6
3 D5

B1 1.09518168
B2 1.09518168
B3 1.09518168
B4 1.53011970
B5 1.09518168
B6 1.09518168
B7 1.09518168
A1 107.49430338
A2 107.49430328
A3 111.38236537
A4 111.38236533
A5 111.38236537
A6 111.38236537
D1 115.45609965
D2 122.27195017
D3 -59.99999997 23 15.0
D4 60.00000006
D5 -180.00000000

1 2 1.0 3 1.0 4 1.0 5 1.0
2
3
4
5 6 1.0 7 1.0 8 1.0
6
7
8

As you can see im trying to rotate the torsion D3 by 15 degrees 23 times.

However Gaussian always crashs this is the message in the log file.

8 H 1.766342 1.766342 0.000000
Small interatomic distances encountered:
7 6
Problem with the distance matrix.
Error termination via Lnk1e in /opt/gaussian03/g03/l202.exe at Mon Mar
6 11:46:01 2006.
Job cpu time: 0 days 0 hours 1 minutes 9.9 seconds.
File lengths (MBytes): RWF= 24 Int= 0 D2E= 0 Chk= 18
Scr= 1

Now i suppose this is pretty self explainatory. But how do i fix it ? im
using B3LYP/6-31(d,p) because a paper I found was also using it.

I found this example online.

http://amber.scripps.edu/Questions/mail/285.html

as far as i can see my .com file is pretty similar.

Does any one have anyother tips for scanning ?

Thanks

Roger






--0-629923880-1141645127=:45467-- From owner-chemistry@ccl.net Mon Mar 6 10:24:00 2006 From: "Marius Retegan marius.retegan,+,yahoo.com" To: CCL Subject: CCL:G: [CCL]Re: CCL:G: Gaussian Scan Message-Id: <-31123-060306075032-32183-kQaPmukqzVRHC1ZkulcnKg,server.ccl.net> X-Original-From: Marius Retegan Content-Transfer-Encoding: 8bit Content-Type: multipart/alternative; boundary="0-1134149394-1141649429=:65444" Date: Mon, 6 Mar 2006 04:50:29 -0800 (PST) MIME-Version: 1.0 Sent to CCL by: Marius Retegan [marius.retegan()yahoo.com] --0-1134149394-1141649429=:65444 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: 8bit Dear Roger I think that your problem is a missing "s", which stands for scan in the input. D3 -59.99999997 23 15.0, should be D3 -59.99999997 s 23 15.0. A good place to learn about scans and geometry optimizations is the page of Prof. Dr. Hendrik Zipse. http://www.cup.uni-muenchen.de/oc/zipse/compchem/topics.html I hope this helps. Marius "Roger Kevin Robinson r.robinson : imperial.ac.uk" wrote: Sent to CCL by: Roger Kevin Robinson [r.robinson]~[imperial.ac.uk] Hi, Im trying to learn how to use the gaussian scan feature. I decided to start with rotating ethane. This is my input file. %chk=/home2/rkr79/gaussian/scans/C2H6-scan.chk %mem=6MW %nproc=1 # scan rb3lyp/6-31g(d,p) geom=connectivity nosym Title Card Required 0 1 C H 1 B1 H 1 B2 2 A1 H 1 B3 3 A2 2 D1 C 1 B4 3 A3 4 D2 H 5 B5 1 A4 3 D3 H 5 B6 1 A5 3 D4 H 5 B7 1 A6 3 D5 B1 1.09518168 B2 1.09518168 B3 1.09518168 B4 1.53011970 B5 1.09518168 B6 1.09518168 B7 1.09518168 A1 107.49430338 A2 107.49430328 A3 111.38236537 A4 111.38236533 A5 111.38236537 A6 111.38236537 D1 115.45609965 D2 122.27195017 D3 -59.99999997 23 15.0 D4 60.00000006 D5 -180.00000000 1 2 1.0 3 1.0 4 1.0 5 1.0 2 3 4 5 6 1.0 7 1.0 8 1.0 6 7 8 As you can see im trying to rotate the torsion D3 by 15 degrees 23 times. However Gaussian always crashs this is the message in the log file. 8 H 1.766342 1.766342 0.000000 Small interatomic distances encountered: 7 6 Problem with the distance matrix. Error termination via Lnk1e in /opt/gaussian03/g03/l202.exe at Mon Mar 6 11:46:01 2006. Job cpu time: 0 days 0 hours 1 minutes 9.9 seconds. File lengths (MBytes): RWF= 24 Int= 0 D2E= 0 Chk= 18 Scr= 1 Now i suppose this is pretty self explainatory. But how do i fix it ? im using B3LYP/6-31(d,p) because a paper I found was also using it. I found this example online. http://amber.scripps.edu/Questions/mail/285.html as far as i can see my .com file is pretty similar. Does any one have anyother tips for scanning ? Thanks Rogerhttp://www.ccl.net/cgi-bin/ccl/send_ccl_messagehttp://www.ccl.net/chemistry/sub_unsub.shtmlhttp://www.ccl.net/spammers.txt--0-1134149394-1141649429=:65444 Content-Type: text/html; charset=iso-8859-1 Content-Transfer-Encoding: 8bit Dear Roger

I think that your problem is a missing "s", which stands for scan in the input.
D3 -59.99999997 23 15.0, should be
D3 -59.99999997 s 23 15.0.
A good place to learn about scans and geometry optimizations is the page of  Prof. Dr.  Hendrik Zipse.
http://www.cup.uni-muenchen.de/oc/zipse/compchem/topics.html
I hope this helps.
Marius

"Roger Kevin Robinson r.robinson : imperial.ac.uk" <owner-chemistry],[ccl.net> wrote:
Sent to CCL by: Roger Kevin Robinson [r.robinson]~[imperial.ac.uk]
Hi,

Im trying to learn how to use the gaussian scan feature. I decided
to start with rotating ethane.
This is my input file.

%chk=/home2/rkr79/gaussian/scans/C2H6-scan.chk
%mem=6MW
%nproc=1
# scan rb3lyp/6-31g(d,p) geom=connectivity nosym

Title Card Required

0 1
C
H 1 B1
H 1 B2 2 A1
H 1 B3 3 A2
2 D1
C 1 B4 3 A3
4 D2
H 5 B5 1 A4
3 D3
H 5 B6 1 A5
3 D4
H 5 B7 1 A6
3 D5

B1 1.09518168
B2 1.09518168
B3 1.09518168
B4 1.53011970
B5 1.09518168
B6 1.09518168
B7 1.09518168
A1 107.49430338
A2 107.49430328
A3 111.38236537
A4 111.38236533
A5 111.38236537
A6 111.38236537
D1 115.45609965
D2 122.27195017
D3 -59.99999997 23 15.0
D4 60.00000006
D5 -180.00000000

1 2 1.0 3 1.0 4 1.0 5 1.0
2
3
4
5 6 1.0 7 1.0 8 1.0
6
7
8

As you can see im trying to rotate the torsion D3 by 15 degrees 23 times.

However Gaussian always crashs this is the message in the log file.

8 H 1.766342 1.766342 0.000000
Small interatomic distances encountered:
7 6
Problem with the distance matrix.
Error termination via Lnk1e in /opt/gaussian03/g03/l202.exe at Mon Mar
6 11:46:01 2006.
Job cpu time: 0 days 0 hours 1 minutes 9.9 seconds.
File lengths (MBytes): RWF= 24 Int= 0 D2E= 0 Chk= 18
Scr= 1

Now i suppose this is pretty self explainatory. But how do i fix it ? im
using B3LYP/6-31(d,p) because a paper I found was also using it.

I found this example online.

http://amber.scripps.edu/Questions/mail/285.html

as far as i can see my .com file is pretty similar.

Does any one have anyother tips for scanning ?

Thanks

Roger






--0-1134149394-1141649429=:65444-- From owner-chemistry@ccl.net Mon Mar 6 10:58:03 2006 From: "Tarmo Tamm tarmo---chem.ut.ee" To: CCL Subject: CCL: Gaussian Scan Message-Id: <-31124-060306102303-7786-a4e1RSOr9LFRD2BhHkgQbg-*-server.ccl.net> X-Original-From: Tarmo Tamm Content-Disposition: inline Content-Type: text/plain; charset=us-ascii Date: Mon, 6 Mar 2006 17:22:54 +0200 Mime-Version: 1.0 Sent to CCL by: Tarmo Tamm [tarmo#%#chem.ut.ee] Hi Opt will initiate a relaxed scan, the "scan" keyword does a restricted scan, these are two different things, consult the manual. You need to define the rotatable group in a way that no dihedral is fixed in place by the hydrogens of the other group. Carbon is ok, as it is placed on the axis of rotation. the following z-matrix will do the trick: c c 1 cc2 h 2 hc3 1 hcc3 h 2 hc4 3 hch4 1 dih4 h 2 hc5 4 hch5 1 dih5 h 1 hc6 2 hcc6 4 dih6 h 1 hc7 6 hch7 2 dih7 h 1 hc8 6 hch8 7 dih8 cc2 1.450000 hc3 1.089000 hcc3 109.471 hc4 1.089000 hch4 109.471 dih4 109.500 hc5 1.089000 hch5 109.471 dih5 109.000 hc6 1.089000 hcc6 109.471 dih6 0.000 35 10.0 hc7 1.089000 hch7 109.471 dih7 109.000 hc8 1.089000 hch8 109.471 dih8 120.000 A rotation of 35 times 10 deg each (it will also calculate the 1st point) Hope it helps Tarmo Tamm University of Tartu On Mar 06, Roger Kevin Robinson r.robinson++imperial.ac.uk wrote: > Sent to CCL by: Roger Kevin Robinson [r.robinson : imperial.ac.uk] > Hi i tried setting each torsion individualy but unsprisingly it did this > > 16 -60.000000 23 15.0000 > 17 60.000000 23 15.0000 > 18 -180.000000 23 15.0000 > A total of 13824 points will be computed. > > which is nt what i want at all and the same problem is still there. > > Do I need to use this keyword > > opt=z-matrix instead of scan > > I have tried this and runs through but it seems to calculate more energy > values than i was expecting and it doesnt put them in a neat little > table like scan does. > > Any reason why it would do this ? > > Is there any way of telling a whole group to rotate ? > > Roger > > From owner-chemistry@ccl.net Mon Mar 6 12:10:01 2006 From: "Roger Kevin Robinson r.robinson(0)imperial.ac.uk" To: CCL Subject: CCL: Gaussian Scan Message-Id: <-31125-060306120030-32464-fyXFECXz5nRVeFJFGyZ6mQ\a/server.ccl.net> X-Original-From: Roger Kevin Robinson Content-Transfer-Encoding: 7bit Content-Type: text/plain; charset=ISO-8859-1; format=flowed Date: Mon, 06 Mar 2006 17:05:36 +0000 MIME-Version: 1.0 Sent to CCL by: Roger Kevin Robinson [r.robinson]=[imperial.ac.uk] Just wanted to say thanks to every one who replied, I think I've got a handle on this now. Roger From owner-chemistry@ccl.net Mon Mar 6 12:44:01 2006 From: "Joaquin Barroso Flores joaco_barroso]![yahoo.com" To: CCL Subject: CCL: Gaussian Scan Message-Id: <-31126-060306121314-4441-iw4K6o9Fx6wW5usy2sP+zA###server.ccl.net> X-Original-From: Joaquin Barroso Flores Content-Transfer-Encoding: 8bit Content-Type: multipart/alternative; boundary="0-1082529302-1141661586=:60077" Date: Mon, 6 Mar 2006 10:13:06 -0600 (CST) MIME-Version: 1.0 Sent to CCL by: Joaquin Barroso Flores [joaco_barroso(!)yahoo.com] --0-1082529302-1141661586=:60077 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: 8bit Hi, you have to define all dihedral angles associated with the torsion you are trying to do, in this case the remainning two C-H bonds in the methyl residue you want to move. After your input coordinates type * 2 3 * 1 2 3 4 S 23 15.0 -blank line- (considering that 2 and 3 are the C atoms) the first line will make all dihedral angles associated with this bond to behave the same way. The second line specifies one of those angles (arbitrarily chosen if you want), then scan (as stated by Marius Retegan on this thread) the number of steps and the increment in each step. Im almost sure you have to use also Opt=Modredundant with this but you may try with and without :) Hope this helps, best regards ********************************************************** Q. Joaquin Barroso-Flores Departamento de Quimica Teorica Instituto de Quimica UNAM Correo Alterno: joaquin_barroso+*+correo.unam.mx joaquin.barroso+*+gmail.com ********************************************************** --------------------------------- Do You Yahoo!? La mejor conexión a Internet y 2GB extra a tu correo por $100 al mes. http://net.yahoo.com.mx --0-1082529302-1141661586=:60077 Content-Type: text/html; charset=iso-8859-1 Content-Transfer-Encoding: 8bit
Hi,
 
you have to define all dihedral angles associated with the torsion you are trying to do, in this case the remainning two C-H bonds in the methyl residue you want to move. After your input coordinates type
 
* 2  3 *
1 2 3 4 S 23 15.0
-blank line-
 
(considering that 2 and 3 are the C atoms) the first line will make all dihedral angles associated with this bond to behave the same way. The second line specifies one of those angles (arbitrarily chosen if you want), then scan (as stated by Marius Retegan on this thread) the number of steps and the increment in each step. Im almost sure you have to use also Opt=Modredundant with this but you may try with and without :)
 
Hope this helps, best regards
 



**********************************************************
Q. Joaquin Barroso-Flores

Departamento de Quimica Teorica
Instituto de Quimica UNAM

Correo Alterno: joaquin_barroso+*+correo.unam.mx
joaquin.barroso+*+gmail.com

**********************************************************

Do You Yahoo!? La mejor conexión a Internet y 2GB extra a tu correo por $100 al mes. http://net.yahoo.com.mx --0-1082529302-1141661586=:60077-- From owner-chemistry@ccl.net Mon Mar 6 13:19:01 2006 From: "Joao Brandao jbrandao-.-ualg.pt" To: CCL Subject: CCL: Thermodynamic reversibility criterion Message-Id: <-31127-060306122638-19298-nE+s92RAwqyFfgns+Twqow{:}server.ccl.net> X-Original-From: "Joao Brandao" Content-Transfer-Encoding: 7bit Content-Type: text/plain; format=flowed; charset="iso-8859-1"; reply-type=original Date: Mon, 6 Mar 2006 16:23:16 -0000 MIME-Version: 1.0 Sent to CCL by: "Joao Brandao" [jbrandao##ualg.pt] Dear CCL members To my knowledge there is some misunderstanding of thermodynamics reversibility in this question. As I understand, in classical thermodynamics there are states (characterized by fixed values of the, so called, state functions) and processes that link two different states (characterized by different pathways). Only processes are reversible or irreversible, depending on the passway. A system in chemical equilibrium at fixed conditions is in a defined STATE of the system and cannot be classified as reversible or irreversible. If we change the system conditions (e.g., raising the total pressure) the system will evolve to a different equilibrium state. Is the way the change is made that can be reversible or not. The reversible way should be a hypothetical way where the raise in pressure is so slow that the system is able to go from one state of equilibrium to the final state through an infinite series of equilibriums. In that pathway, the TOTAL (system and surroundings) change of entropy should be zero. The change of entropy of the system depends only on the initial and final states, it is independent of the way chosen to increase pressure, and can be different from zero. This classical thermodynamics definition of reversibility is not related with direct and reverse reactions (and their rate constants) that should occur in all equilibriums. Of course, it is easier to devise a reversible pathway if both reactions are fast, but this is not thermodynamics is a problem of kinetics. Also, this definition it is not related with the well known microscopic reversibility of the equations of motion. Joao Brandao Departamento de Quimica e Bioquimica Faculdade de Ciencias e Tecnologia Universidade do Algarve - Campus de Gambelas P 8005-139 FARO Portugal www: http://w3.ualg.pt/~jbrandao e-mail: jbrandao!A!ualg.pt telem: 918465718 ----- Original Message ----- > From: "Ramon Crehuet rcsqtc=iiqab.csic.es" To: "Brandao, Joao " Sent: Friday, March 03, 2006 3:19 PM Subject: CCL: Thermodynamic reversibility criterion > Sent to CCL by: Ramon Crehuet [rcsqtc++iiqab.csic.es] > Hi all, > Although microscopic reversibility always holds, even one-step reactions > A-> B can, in practice, be irreversible. This is probably not what you > see in textbooks, but an experimental chemist will probably tell you > that they are irrevesible. This will happen when the products have other > possible reaction paths, apart from that giving back the product A, > specially if the rate of these processes is faster than the rate to give > back the reactant A. So that, in a lab, once you have done A->B if you > change the conditions to reverse the equilibrium, B will give C, D, E... > instead of A. > Cheers, > Ramon > > > > John Simmie john.simmie]_[nuigalway.ie wrote: > >>Sent to CCL by: John Simmie [john.simmie|nuigalway.ie] >> >>--Boundary_(ID_sauQiCYe11HXHeutKYUT0w) >>Content-type: text/plain; charset=us-ascii; format=flowed >> >>Gonzalo: >>All elementary reactions are reversible no matter what the entropy or >>enthalpy changes amount to. >>according to the principle of microscopic reversibility >> >>ie Nature does not discriminate between reactants and products ... >>the paths traced out from >>reactant ==> product are identical to those from product ==> reactant >> >>The connection between kinetics & thermodynamics is given by >>k(forward) / k(reverse) = K >>where k's are rate constants and K is equilibrium constant .... from >>this you can relate >>k(reverse) to k(forward) and Gibbs free energy or enthalpy & entropy >> >>Trust this clarifies the position, John >> >>At 02:58 03/03/2006, you wrote: >> >> >>>Sent to CCL by: "Dr. Seth Olsen" [s.olsen1**uq.edu.au] >>> >>>Hi Gonzalo, >>> >>>A reaction is reversible if the entropy change is zero. If the >>>activation barriers are different for the forward and reverse reactions >>>then the forward and reverse rates will be different, but this is not >>>the same as thermodynamic reversibility. I recommend Dill & Blomberg's >>>'Molecular Driving Forces' for further info, though most chemical >>>thermodynamics texts will do. >>> >>>Cheers, >>> >>>Seth >>> >>> >>> >>>Gonzalo Jimenez Oses gonzalo.jimenez[#]dq.unirioja.es wrote: >>> >>> >>> >>>>Sent to CCL by: "Gonzalo Jimenez Oses" [gonzalo.jimenez a >>>>dq.unirioja.es] >>>>Dear colleagues, >>>> >>>>Could anyone shed some light into the reversivility of chemical >>>> >>>> >>>processes attending to thermodynamic criteria, please?. Everybody >>>tend to say that a reaction is reversible when the activation >>>barrier from the products to a TS is lower or "comparable" with the >>>one which goes from the reactants to the same TS. My question is if >>>there is a Gibbs energy value (or range) that could make us to >>>discriminate between reversible and irreversible processes. >>> >>> >>>>Thank you all> >>>> >>>> >>>> >>>> >>>> >>>> >>>-- >>>ccmsccmsccmsccmsccmsccmsccmsccmsccmsccmsccmsccmsccmsccmsccmsccms >>> >>>Dr Seth Olsen, PhD >>>Postdoctoral Fellow, Biomolecular Modeling Group >>>Centre for Computational Molecular Science >>>Chemistry Building, >>>The University of Queensland >>>Qld 4072, Brisbane, Australia >>> >>>tel (617) 33653732 >>>fax (617) 33654623 >>>email: s.olsen1{:}uq.edu.au >>>Web: www.ccms.uq.edu.au >>> > > > From owner-chemistry@ccl.net Mon Mar 6 13:55:04 2006 From: "Gonzalo Jimenez Oses gonzalo.jimenez**dq.unirioja.es" To: CCL Subject: CCL: Thermodynamic reversibility criterion - Summary - Message-Id: <-31128-060306125602-3282-vbSareQRVtiW0lqYll/27A_._server.ccl.net> X-Original-From: "Gonzalo Jimenez Oses" Date: Mon, 6 Mar 2006 12:56:02 -0500 Sent to CCL by: "Gonzalo Jimenez Oses" [gonzalo.jimenez-.-dq.unirioja.es] Dear CCL'ers: I would like to thank for all the quick and clarifying answers I have recieved concerning my question. Here is a summary with my quesstion and all the replies. Good luck! Gonzalo ------------------------------------------------------------------------- Sent to CCL by: "Gonzalo Jimenez Oses" [gonzalo.jimenez a dq.unirioja.es] Dear colleagues, Could anyone shed some light into the reversivility of chemical processes attending to thermodynamic criteria, please?. Everybody tend to say that a reaction is reversible when the activation barrier from the products to a TS is lower or "comparable" with the one which goes from the reactants to the same TS. My question is if there is a Gibbs energy value (or range) that could make us to discriminate between reversible and irreversible processes. Thank you all ------------------------------------------------------------------------- Sent to CCL by: "Dr. Seth Olsen" [s.olsen1**uq.edu.au] Hi Gonzalo, A reaction is reversible if the entropy change is zero. If the activation barriers are different for the forward and reverse reactions then the forward and reverse rates will be different, but this is not the same as thermodynamic reversibility. I recommend Dill & Blomberg's 'Molecular Driving Forces' for further info, though most chemical thermodynamics texts will do. Cheers, Seth ------------------------------------------------------------------------- Sent to CCL by: John Simmie [john.simmie|nuigalway.ie] Gonzalo: All elementary reactions are reversible no matter what the entropy or enthalpy changes amount to. according to the principle of microscopic reversibility ie Nature does not discriminate between reactants and products ... the paths traced out from reactant ==> product are identical to those from product ==> reactant The connection between kinetics & thermodynamics is given by k(forward) / k(reverse) = K where k's are rate constants and K is equilibrium constant .... from this you can relate k(reverse) to k(forward) and Gibbs free energy or enthalpy & entropy Trust this clarifies the position, John ------------------------------------------------------------------------- Sent to CCL by: "Mariusz Radon" [mariusz.radon++gmail.com] Hi, In general case TdS-DQ = 0 (where DQ is a differential form of heat) is a condition for thermodynamic equilibrium (therm. reversibility). This equation is a basis for deriving well known expression under different conditions, such as dG=0 when p,T=const. It is important that thermodynamic reversibitly is not a feature of reaction (i.e. is determined not only by reagents, but depend also on conditions and reaction progress). So, I think you asked about chemical reversibility. If we go into thermodynamics considerations we ussually assume state of equilibrium (th. reversibility) has been reached. Restricting to isobaric and isotermic processes and taking into account the equation dG=0, where G depends on concentration of reagents, one may derive Guldberg-Waage law, with chemical equilibrium constant, K. In this sense each chemical process *is* reversible (because both "subst->prod" like "prod->subst" reactions are taking part), but if K is very huge we sometimes call it "irreversible". This means that in the state of equilibrium the concentration of products is much bigger than substrates and the velocity of forming products exceed the velocity of re-forming substrates. Because in mentioned conditions (p,T=const): K=\exp(-\Delta G^0 / RT) one may conclude that reaction is practically irreversible (in chem. sense!) when \Delta G^0 << 0. Here \Delta G^0 means "standard change of Gibb's free enalphy", which is the difference between the sum of G (per mol) for products and substrates (not dG in thermodynamics!). In some practical cases G \approx E + const, thus the criterium given by Gonzalo holds. But of course sometimes changes of entrophy and volume comes into play -- than one should employ full Gibb's free entalphy. Take care, Mariusz ------------------------------------------------------------------------- Sent to CCL by: Thomas H Dr Pierce [TPierce++rohmhaas.com] Gonzalo, Another generalization is that all physical processes are irreversible. The reason we do not have any perpetual motion systems is that a reaction loses some entropy or heat. If we define the "system" to allow for modest energy back into the system, we get reactions that behave as if they are reversible. But they consume energy so the are not strictly thermodynamically reversible. I consider the principle of microscopic reversibility to mean that a path in either direction is possible given the energy input or output needed. Water breaking into ions and back comes to mind as a room temperature event. The rate of forward to backward reactions assumes an energy source or bath. High transition barriers can be overcome if there is sufficient energy. Making Diamond from carbon and of course burning diamond is an example of this at high barriers requiring significant energy. It always seems like it is the definition of the "system" that carries the answer. Sincerely, Tom Pierce ------------------------------------------------------------------------- Sent to CCL by: "Lukasz Cwiklik" [cwiklik]^[gmail.com] Dear Gonzalo, In my opinion (I hope my remarks are not too general), the ratio of values of activation barriers on both sides of TS lets you to calculate the populations (or probabilities of observing, or cancentrations) of both reactatants and products (but only if your system is in thermodynamical equilibrium). So taking the Gibbs energy values you can calculate probabilities/rates of processes but you cannot _strictly_ discriminate between reversibility and irreversibility. One thing that is also important is the possibility of other processes in your system. If your reactants may form a product which then may undergo the reverse transition _or_ react along some other path, then you must be careful and, in your considerations of reversibility, you should take into account the barrier of this third process. Best regards, Lukasz ------------------------------------------------------------------------- hi Gonzalo... The words 'reversible' and 'irreversible' are used to mean two rather unrelated things... First, the words are used to describe the thermodynamics of processes. A reversible process is one for which the net change of the entropy of the universe is 0. Or equivalently, the change in helmholtz free energy of the system is 0 if kept at constant volume and temperature, or the change in the gibbs free energy of the system is 0 if kept at constant pressure and temperature. Of course, such a process is an idealization and can never be achieved in practice. Any real process is irreversible, meaning that the net change of the entropy of the universe is > 0 (or equivalently, the change in the helmholtz free energy of the system is < 0 if kept at constant volume, etc.) Second, the words are used to describe the kinetics of chemical reactions. A reversible chemical reaction is one for which the activation barrier (Helmholtz free energy difference between transition state and reactant if kept at constant volume, Gibbs free energy difference... if kept at constant pressure) is not enormously greater than kT. Such a reaction reaches thermodynamic equilibrium on a reasonable timescale. An example would be the autoionization of water. An irreversible chemical reaction is one for which the activation barrier is much larger than kT. An example would be diamond -> graphite, which is thermodynamically favored, but does not reach thermodynamic equilibrium at room temperature on any but geologic time scales, thankfully for de Beers. Hope that helps Harry ------------------------------------------------------------------------- Sent to CCL by: Ramon Crehuet [rcsqtc++iiqab.csic.es] Hi all, Although microscopic reversibility always holds, even one-step reactions A-> B can, in practice, be irreversible. This is probably not what you see in textbooks, but an experimental chemist will probably tell you that they are irrevesible. This will happen when the products have other possible reaction paths, apart from that giving back the product A, specially if the rate of these processes is faster than the rate to give back the reactant A. So that, in a lab, once you have done A->B if you change the conditions to reverse the equilibrium, B will give C, D, E... instead of A. Cheers, Ramon From owner-chemistry@ccl.net Mon Mar 6 14:29:01 2006 From: "Deepangi Pandit deepangi.pandit_+_gmail.com" To: CCL Subject: CCL:G: Gaussian Scan Message-Id: <-31129-060306114232-27139-z0lVAhkLMUy769J+MWtl3g#%#server.ccl.net> X-Original-From: "Deepangi Pandit" Content-Disposition: inline Content-Transfer-Encoding: 8bit Content-Type: text/plain; charset=ISO-8859-1 Date: Mon, 6 Mar 2006 10:47:14 -0500 MIME-Version: 1.0 Sent to CCL by: "Deepangi Pandit" [deepangi.pandit(0)gmail.com] Yes opt=z-matrix should be used for scan. I use following command to use scan %chk= filename.chk %mem=128MW %nproc=1 # opt=z-matrix hf/6-31g(d) nosymm and I put s near dihedral which I want to rotate. In your case you should put this on the dihedral which defines C2, which you want to rotate. It seems to me it is D2 in your case. The below is just for example d2 60.0000 s 13 30.0 Deepangi On 3/6/06, Roger Kevin Robinson r.robinson++imperial.ac.uk wrote: > Sent to CCL by: Roger Kevin Robinson [r.robinson : imperial.ac.uk] > Hi i tried setting each torsion individualy but unsprisingly it did this > > 16 -60.000000 23 15.0000 > 17 60.000000 23 15.0000 > 18 -180.000000 23 15.0000 > A total of 13824 points will be computed. > > which is nt what i want at all and the same problem is still there. > > Do I need to use this keyword > > opt=z-matrix instead of scan > > I have tried this and runs through but it seems to calculate more energy > values than i was expecting and it doesnt put them in a neat little > table like scan does. > > Any reason why it would do this ? > > Is there any way of telling a whole group to rotate ? > > Roger > > > Tarmo Tamm tarmo[]chem.ut.ee wrote: > > >Sent to CCL by: Tarmo Tamm [tarmo[]chem.ut.ee] > >Hi > > > >Looks like you are just rotating one hydrogen instead of the CH3 group. > >You should define the remaining 2 hydrogens depending on the one you are > >rotating, in order to make them move together not towards each other. > > > >Good luck > > > >Tarmo Tamm > >University of Tartu > > > > > > > >On Mar 06, Roger Kevin Robinson r.robinson : imperial.ac.uk wrote: > > > > > >>Sent to CCL by: Roger Kevin Robinson [r.robinson]~[imperial.ac.uk] > >>Hi, > >> > >> Im trying to learn how to use the gaussian scan feature. I decided > >>to start with rotating ethane. > >>This is my input file.> > > > From owner-chemistry@ccl.net Mon Mar 6 15:04:00 2006 From: "Dimitrios Pantazis pantazis:-:hotmail.com" To: CCL Subject: CCL:G: Gaussian Scan Message-Id: <-31130-060306115504-29938-R0uJMP4+b0vJ5UAmDkAhgw{}server.ccl.net> X-Original-From: "Dimitrios Pantazis" Content-Transfer-Encoding: 7bit Content-Type: text/plain; format=flowed; charset="iso-8859-1"; reply-type=original Date: Mon, 6 Mar 2006 16:08:33 -0000 MIME-Version: 1.0 Sent to CCL by: "Dimitrios Pantazis" [pantazis-x-hotmail.com] Hi Roger, as pointed out already, the problem is caused by the specific way that your z-matrix is constructed, i.e. the last two H atoms (7 and 8) are independent of the 6th hydrogen, so they do not move as the dihedral D3 is stepped. To correct that, you should define these hydrogens using the dihedrals with respect to the rotating hydrogen. The last two lines in your z-matrix should be: .. H 5 B6 1 A5 6 D4 H 5 B7 1 A6 6 D5 where the sixth atom (instead of the third) is used to define dihedrals D4 and D5. The values of D4 and D5 should of course be changed as well: .. D4 -120.0 D5 120.0 Everything else in your input should be OK. Dimitrios --- Dr. Dimitrios A. Pantazis Computational Chemistry Group WestCHEM Department of Chemistry Joseph Black Building University of Glasgow Glasgow G12 8QQ, UK --- > Sent to CCL by: Marius Retegan [marius.retegan^^yahoo.com] > --0-629923880-1141645127=:45467 > Content-Type: text/plain; charset=iso-8859-1 > Content-Transfer-Encoding: 8bit > > Dear Roger > > I think that your problem is a missing "s", which stands for scan in the > input. > D3 -59.99999997 23 15.0, should be > D3 -59.99999997 s 23 15.0. > A good place to learn about scans and geometry optimizations is the page > of Prof. Dr. Hendrik Zipse. > http://www.cup.uni-muenchen.de/oc/zipse/compchem/topics.html > I hope this helps. > Marius > "Roger Kevin Robinson r.robinson : imperial.ac.uk" > wrote: Sent to CCL by: Roger Kevin Robinson > [r.robinson]~[imperial.ac.uk] > Hi, > > Im trying to learn how to use the gaussian scan feature. I decided > to start with rotating ethane. > This is my input file. > > %chk=/home2/rkr79/gaussian/scans/C2H6-scan.chk > %mem=6MW > %nproc=1 > # scan rb3lyp/6-31g(d,p) geom=connectivity nosym > > Title Card Required > > 0 1 > C > H 1 B1 > H 1 B2 2 A1 > H 1 B3 3 A2 > 2 D1 > C 1 B4 3 A3 > 4 D2 > H 5 B5 1 A4 > 3 D3 > H 5 B6 1 A5 > 3 D4 > H 5 B7 1 A6 > 3 D5 > > B1 1.09518168 > B2 1.09518168 > B3 1.09518168 > B4 1.53011970 > B5 1.09518168 > B6 1.09518168 > B7 1.09518168 > A1 107.49430338 > A2 107.49430328 > A3 111.38236537 > A4 111.38236533 > A5 111.38236537 > A6 111.38236537 > D1 115.45609965 > D2 122.27195017 > D3 -59.99999997 23 15.0 > D4 60.00000006 > D5 -180.00000000 > > 1 2 1.0 3 1.0 4 1.0 5 1.0 > 2 > 3 > 4 > 5 6 1.0 7 1.0 8 1.0 > 6 > 7 > 8 > > As you can see im trying to rotate the torsion D3 by 15 degrees 23 times. > > However Gaussian always crashs this is the message in the log file. > > 8 H 1.766342 1.766342 0.000000 > Small interatomic distances encountered: > 7 6 > Problem with the distance matrix. > Error termination via Lnk1e in /opt/gaussian03/g03/l202.exe at Mon Mar > 6 11:46:01 2006. > Job cpu time: 0 days 0 hours 1 minutes 9.9 seconds. > File lengths (MBytes): RWF= 24 Int= 0 D2E= 0 Chk= 18 > Scr= 1 > > Now i suppose this is pretty self explainatory. But how do i fix it ? im > using B3LYP/6-31(d,p) because a paper I found was also using it. > > I found this example online. > > http://amber.scripps.edu/Questions/mail/285.html > > as far as i can see my .com file is pretty similar. > > Does any one have anyother tips for scanning ? > > Thanks > > Rogerhttp://www.ccl.net/cgi-bin/ccl/send_ccl_messagehttp://www.ccl.net/chemistry/sub_unsub.shtmlhttp://www.ccl.net/spammers.txt--0-629923880-1141645127=:45467 > Content-Type: text/html; charset=iso-8859-1 > Content-Transfer-Encoding: 8bit > > Dear Roger

I think that your problem is a missing "s", which stands > for scan in the input.
D3 -59.99999997 23 15.0, should > be
D3 -59.99999997 s 23 15.0.
A good place to learn > about scans and geometry optimizations is the page of  Prof. > Dr.  Hendrik > Zipse.
http://www.cup.uni-muenchen.de/oc/zipse/compchem/topics.html
I > hope this helps.
Marius
"Roger Kevin Robinson r.robinson : > imperial.ac.uk" <owner-chemistry.:.ccl.net> > wrote:
Sent to CCL by: Roger Kevin > Robinson [r.robinson]~[imperial.ac.uk]
Hi,

Im trying to > learn how to use the gaussian scan feature. I decided
to start with > rotating ethane.
This is my input > file.

%chk=/home2/rkr79/gaussian/scans/C2H6-scan.chk
%mem=6MW
%nproc=1
# > scan rb3lyp/6-31g(d,p) geom=connectivity nosym

Title Card > Required

0 1
C
H 1 B1
H > 1 B2 2 A1
H 1 > B3 3 A2
2 D1
C 1 > B4 3 A3
4 D2
H 5 > B5 1 A4
3 D3
H 5 > B6 1 A5
3 D4
H 5 > B7 1 A6
3 D5

B1 > 1.09518168
B2 1.09518168
B3 > 1.09518168
B4 1.53011970
B5 > 1.09518168
B6 1.09518168
B7 > 1.09518168
A1 107.49430338
A2 > 107.49430328
A3 111.38236537
A4 > 111.38236533
A5 111.38236537
A6 > 111.38236537
! > D1 > 115.45609965
D2 122.27195017
> -59.99999997 23 15.0
D4 60.00000006
> -180.00000000

1 2 1.0 3 1.0 4 1.0 5 1.0
2
3
4
> 5 6 1.0 7 1.0 8 1.0
6
7
8

As you can see im trying to > rotate the torsion D3 by 15 degrees 23 times.

However Gaussian > always crashs this is the message in the log file.

8 H > 1.766342 1.766342 0.000000
Small interatomic distances > encountered:
7 6
Problem with the distance matrix.
> Error termination via Lnk1e in /opt/gaussian03/g03/l202.exe at Mon Mar >
6 11:46:01 2006.
Job cpu time: 0 days 0 hours 1 minutes 9.9 > seconds.
File lengths (MBytes): RWF= 24 Int= 0 D2E= 0 > Chk= 18
Scr= 1

Now i suppose this is pretty self > explainatory. But how do i fix it ? im
using B3LYP/6-31(d,p) because a > paper I found was also using it.

I found this example > online.

http://amber.scripps.edu/Questions/mail/285.html

as > far as i can see my .com file is pretty similar.

Does any one have > anyother tips for scanning > ?

Thanks

Roger






> --0-629923880-1141645127=:45467--> > > > From owner-chemistry@ccl.net Mon Mar 6 15:39:01 2006 From: "Shobe, David dshobe^^^sud-chemieinc.com" To: CCL Subject: CCL:G: Transition state for recombination of dibromomercury Message-Id: <-31131-060306115522-30013-ADPZSnZBydNLOQx1ls+izg]|[server.ccl.net> X-Original-From: "Shobe, David" Content-class: urn:content-classes:message Content-Transfer-Encoding: 8bit Content-Type: text/plain; charset="iso-8859-1" Date: Mon, 6 Mar 2006 17:55:12 +0100 MIME-Version: 1.0 Sent to CCL by: "Shobe, David" [dshobe:_:sud-chemieinc.com] Sent to CCL by: "luis simon" [luissimonrubio:_:hotmail.com] > Although I am convinced by David arguments about the unnecesity of a TS, > I keep still a doubt: how can QM method it distinguish between the homolitic > (i.e. diradical) and the heterolytic (anion/cation) reaction. That certainly can be a problem..."restricted" methods CAN force a heterolytic dissociation even when the homolytic dissociation is lower in energy. But you can use unrestricted methods for a singlet calculation also, you just have to explicitly specify this in the job file. In Gaussian you put a "u" in front of the method keyword: "ump2" or "ub3lyp" for example. So I recommend that you calculate the PES using an unrestricted method, and look for sudden jumps in energy which can indicate a jump from the homolytic to the heterolytic PES. Normally I'd say the singlet and triplet PES don't have anything to do with each other, but BrHg...Br could have very significant spin-orbit coupling so who knows? It's beyond my expertise anyway. --David Shobe, Ph.D., M.L.S. Süd-Chemie, Inc. phone (502) 634-7409 fax (502) 634-7724 Don't bother flaming me: I'm behind a firewall. -----Original Message----- > From: owner-chemistry,,ccl.net [mailto:owner-chemistry,,ccl.net] Sent: Saturday, February 25, 2006 1:59 PM To: Shobe, David Subject: CCL:G: Transition state for recombination of dibromomercury Sent to CCL by: "luis simon" [luissimonrubio:_:hotmail.com] After reading the message from David Shobe I am not pretty sure if I did right changing the multiplicity to 3 in a diradical reaction. My first attempts were to set multiplicity to 1 in the transition state search, but as this did not produce any result, I considered to change it to 3 (although it lacks logic for me) and I obtained a converged transition state with one imaginary frequency corresponding to the bond that was forming in the reaction. The level of theory that I was using was certainly poor: UHF/3-21G, but I considered that the neccesity for multiplicity 3 was some artifact of the unrestristed HF. Although I am convinced by David arguments about the unnecesity of a TS, I keep still a doubt: how can QM method it distinguish between the homolitic (i.e. diradical) and the heterolytic (anion/cation) reaction. Does it rely exclusivelly on the unrestristed/restristed HF method? I am really sorry for posting the "trick" that I found now that I consider that probably I was wrong, and I specially apologize to Ian Hovell if I was terribly wrong. But could anyone please answer my former question? Thanks: Luis ----- Original Message ----- > From: "Shobe, David dshobe(-)sud-chemieinc.com" ccl.net> To: "Simón, Luis Manuel " Sent: Saturday, February 25, 2006 2:54 AM Subject: CCL:G: Transition state for recombination of dibromomercury > Sent to CCL by: "Shobe, David" [dshobe,sud-chemieinc.com] Ian, > > Radical recombination reactions often don't have a TS, and it's not unusual to have a TS at one level of theory and no TS at another. > > My advice is to run a PES (potential energy scan) varying the Hg-Br distance. If the energy is a monotonic function of the bond length, then there is no transition state (according to whichever level of theory you are using). > > --David Shobe, Ph.D., M.L.S. > Süd-Chemie, Inc. > phone (502) 634-7409 > fax (502) 634-7724 > > Don't bother flaming me: I'm behind a firewall. > > > > -----Original Message----- > > From: owner-chemistry|ccl.net [mailto:owner-chemistry|ccl.net] > Sent: Friday, February 24, 2006 2:44 PM > To: Shobe, David > Subject: CCL:G: Transition state for recombination of dibromomercury > > Sent to CCL by: Ian Hovell [HOVELL/./cetem.gov.br] This message is in > MIME format. Since your mail reader does not understand this format, some or all of this message may not be legible. > > ------_=_NextPart_001_01C6393D.C86E54E0 > Content-Type: text/plain; > charset="iso-8859-1" > > Dear CCLers, > I am assuming a linear transition state for the recombination of a > bromine radical with the bromomercury radical forming the dibromo molecule. > According to Nikolai (J Phys. Chem. A 2005, 109, 8765-8773) the two > bond lengths are 2.539 and 4.567 angstroms produced a transition state calculated using CI methods. > I am trying to produce, without much success, a transition state for > this recombination reaction at the mpw1pw91 DFT level. In either full or partial (where only one bond is relaxed) optimisations, the radicals either join forming the dibromo molecule, and hence no imaginary frequency, or separate completly according to the starting bond lengths. > > I am using the built in SDD basis set and G03w. The output produces a warning that the bromine might be hypervalent but has no d functions - could this be the problem or this this level of calculation just not up to finding the transition state for this system? > All ideas would be welcome and a summary will be posted if this post generates interest. > TIA > Ian > > > Ian Hovell - Ph.D. > NUCLEO DE MODELAGEM MOLECULAR-NMM > Centro de Tecnologia Mineral - CETEM > Ministerio da Cincia e da Tecnologia- MCT Avenida Ip, No 900 - Cidade Universitaria Ilha do Fundo Rio de Janeiro RJ Brasil CEP 21941-590 tel 00 55 (xx) 3865 7344 ou 3865 - 7216 Fax 00 55 (xx) 22602837 ou 2290-4286 e-mail hovell###cetem.gov.br > > > > ------_=_NextPart_001_01C6393D.C86E54E0 > Content-Type: text/html; > charset="iso-8859-1" > > > > >
Dear CCLers,
I am assuming a linear transition state for the recombination of a bromine radical with the bromomercury radical forming the dibromo molecule.
According to Nikolai (J Phys. Chem. A 2005, 109, 8765-8773) the two bond lengths are 2.539 and 4.567 angstroms produced a transition state calculated using CI methods.
I am trying to produce, without much success, a transition state for this recombination reaction at the mpw1pw91 DFT level. In either full or partial (where only one bond is relaxed) optimisations, the radicals either join forming the dibromo molecule, and hence no imaginary frequency, or separat! > e completly according to the starting bond > lengths.
 
>
I am using > the built in SDD basis set and G03w. The output produces a warning that the bromine might be hypervalent but has no d functions - could this be the problem or this this level of calculation just not up to finding the transition state for this system?
All ideas would be welcome and a summary will be posted if this post generates interest.
TIA
>
class=718415311-24022006>Ian
>
class=718415311-24022006> 
>
class=718415311-24022006> 
>
Ian Hovell - Ph.D.
NUCLEO DE > MODELAGEM MOLECULAR-NMM
Centro de Tecnologia Mineral - CETEM
Ministerio da Cincia e da Tecnologia- MCT
Avenida Ip, No 900 - Cidade Universitaria
Ilha do Fundo Rio de Janeiro RJ Brasil
CEP 21941-590
tel 00 55 (xx) 3865 7344 ou > 3865 - 7216
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> > ------_=_NextPart_001_01C6393D.C86E54E0--http://www.ccl.net/cgi-bin/ccl/send_ccl_messagehttp://www.ccl.net/chemistry/sub_unsub.shtmlhttp://www.ccl.net/spammers.txtThis e-mail message may contain confidential and / or privileged information. If you are not an addressee or otherwise authorized to receive this message, you should not use, copy, disclose or take any action based on this e-mail or any information contained in the message. If you have received this material in error, please advise the sender immediately by reply e-mail and delete this message. Thank you. From owner-chemistry@ccl.net Mon Mar 6 18:09:00 2006 From: "Fiona Case fhcase%a%hotmail.com" To: CCL Subject: CCL: Predicting useful properties of fluids Message-Id: <-31132-060306172611-24403-np682dmazBtaH+HZ5HzWuw[#]server.ccl.net> X-Original-From: "Fiona Case" Content-Type: text/plain; format=flowed Date: Mon, 06 Mar 2006 16:38:55 -0500 Mime-Version: 1.0 Sent to CCL by: "Fiona Case" [fhcase-.-hotmail.com] Here is an important question: Can molecular simulations methods be used to predict industrially relevant properties of fluids (such as viscosity, vapor pressure, heat of vaporization, boiling temperature, heat capacity, isothermal compressibility, thermal conductivity, vapor liquid equilibria, or infinite dilution activity coefficients)? How well do the different methods and forcefields perform when compared to experimental data, and to other, non-simulation, prediction techniques? The Industrial Fluid Property Simulation Collective (IFPSC) is a group of chemists and chemical engineers from industry and NIST who want to answer this question! We believe molecular simulation could become a breakthrough technology that is widely accepted in the chemical industry. It has the potential to provide a more cost-effective method for obtaining physical property data than experiments – particularly for materials and state points that are experimentally challenging or hazardous. The IFPSC have already organized two simulation challenges (in 2002 and 2004), and we are about to launch the third. The types of questions we are considering are listed at our website: http://fluidproperties.org/challenge/2006proposals.html The CCL represents a broad cross-section of the computational chemistry community. What do you think - do these questions represent a reasonable test of today's leading methods and forcefields? We ask you to take the time to answer our survey: http://www.surveymonkey.com/s.asp?u=35161747608 While you are at the website take a look at other ideas for expanding the value and acceptance of molecular simulations and other property prediction methods. A vision and strategic plan is available here and your comments are welcome: http://fluidproperties.org/Vision-and-Strategic-Plan-5-Feb-06.pdf Sincerely, The IFPSC organizing committee The IFPSC is an industry-led initiative aimed at promoting and encouraging the development of techniques for the prediction of industrially-relevant fluid properties. The activities of the group are supported by the Computational Molecular Science and Engineering Forum (CoMSEF) of the American Institute of Chemical Engineers (AIChE), the Computational Chemistry (COMP) and Physical Chemistry (PHYS) divisions of the American Chemical Society (ACS), The Dow Chemical Company, BP Amoco, NIST, Case Scientific, ExxonMobil, Mitsubishi Chemical, 3M Company, and DuPont. From owner-chemistry@ccl.net Mon Mar 6 18:44:00 2006 From: "Tianxiao Yang xiaoyang_guelph##yahoo.com" To: CCL Subject: CCL: Multiplicity in GAMESS Message-Id: <-31133-060306154526-15879-kQaPmukqzVRHC1ZkulcnKg . server.ccl.net> X-Original-From: "Tianxiao Yang" Date: Mon, 6 Mar 2006 15:45:25 -0500 Sent to CCL by: "Tianxiao Yang" [xiaoyang_guelph~~yahoo.com] Dear Colleagues, I am confusing with the S and SZ values in GAMESS. When I optimize the ground state (with MULT=1 in the input file) using MCSCF method, I got the following three roots. STATE 1 ENERGY= -1421.2244772662 S= 0.00 SZ= 0.00 SPACE SYM=A STATE 2 ENERGY= -1421.1162694962 S= 1.00 SZ= 0.00 SPACE SYM=A STATE 3 ENERGY= -1421.0552894237 S= 1.00 SZ= 0.00 SPACE SYM=A Is there any way to figure out STATE 2 and STATE 3 Singlet or Triplet? Thank you very much for your kind help. Xiao Yang Department of Chemistry, University of Guelph