From owner-chemistry@ccl.net Sun Aug 29 00:49:01 2010 From: "Grigoriy Zhurko reg_zhurko-.-chemcraftprog.com" To: CCL Subject: CCL:G: Visualization of G09 output Message-Id: <-42631-100829004704-21107-zrAaWIavz1DGymzoFj9dew~~server.ccl.net> X-Original-From: Grigoriy Zhurko Content-Transfer-Encoding: 7bit Content-Type: text/plain; charset=us-ascii Date: Sun, 29 Aug 2010 08:46:56 -0700 MIME-Version: 1.0 Sent to CCL by: Grigoriy Zhurko [reg_zhurko*chemcraftprog.com] > Dear All > I am facing problem in visualizing G09 output. Gaussview (not > reading vibration),Avagadro (not reading optimisation > steps),chemcraft(not reading vibration and optimization steps).These > all were suitable for G03 output.Please suggest me that what should > I do.Is there any free software for visualizing all results. > > Your suggestions will be appreciated and helpful for others too......... If the latest official version of Chemcraft does not read frequencies from G09 outputs, you can download the newest version here: http://www.chemcraftprog.com/files/Chemcraft_b335_win.exe It is not the official release (official version will be released within 1-2 months). > With regards; > ABHISHEK SHAHI > IISc bangalore-12 > Official E-mail: shahi()ipc.iisc.ernet.in > CC: shahi.abhishek1984()gmail.com > Sincerely, Grigoriy Zhurko. From owner-chemistry@ccl.net Sun Aug 29 01:23:00 2010 From: "Farhan Pasha pashafa[*]yahoo.co.in" To: CCL Subject: CCL:G: Visualization of G09 output Message-Id: <-42632-100829011759-15866-/g4lbIUpJUNpwVImvZRD1Q\a/server.ccl.net> X-Original-From: Farhan Pasha Content-Type: multipart/alternative; boundary="0-1210661403-1283059045=:64566" Date: Sun, 29 Aug 2010 10:47:25 +0530 (IST) MIME-Version: 1.0 Sent to CCL by: Farhan Pasha [pashafa^^yahoo.co.in] --0-1210661403-1283059045=:64566 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable Dear Abhishek Try new version of Chemcraft it is working fine for me. Best Pasha Correspondence Address =0ADr. Syed Farhan Ahmad Pasha --- On Sat, 28/8/10, ABHISHEK SHAHI shahi.abhishek1984!^!gmail.com wrote: > From: ABHISHEK SHAHI shahi.abhishek1984!^!gmail.com Subject: CCL:G: Visualization of G09 output To: "Pasha, Frahan Ahmad " Date: Saturday, 28 August, 2010, 9:50 PM Dear All =A0=A0 I am facing problem in visualizing G09 output. Gaussview (not readin= g vibration),Avagadro (not reading optimisation steps),chemcraft(not readin= g vibration and optimization steps).These all were suitable for G03 output.= Please suggest me that what should I do.Is there any free software for visu= alizing all results. =0A =A0Your suggestions will be=A0 appreciated and helpful for=A0 others too...= ...... With regards; =A0ABHISHEK SHAHI =0A=A0 IISc bangalore-12 =A0 Official E-mail: shahi()ipc.iisc.ernet.in =A0 CC:=A0 shahi.abhishek1984()gmail.com =0A=0A=0A=0A --0-1210661403-1283059045=:64566 Content-Type: text/html; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable
Dear Abhishek

Try new version of Chemc= raft it is working fine for me.

Best
Pasha

Correspondence = Address
=0ADr. Syed Farhan Ahmad Pasha

--- On Sat, 28/8/10, AB= HISHEK SHAHI shahi.abhishek1984!^!gmail.com <owner-chemistry _ ccl.net&= gt; wrote:

From: ABHISHEK SHAHI sha= hi.abhishek1984!^!gmail.com <owner-chemistry _ ccl.net>
Subject: CCL= :G: Visualization of G09 output
To: "Pasha, Frahan Ahmad " <p= ashafa _ yahoo.co.in>
Date: Saturday, 28 August, 2010, 9:50 PM

<= div id=3D"yiv1754330165">Dear All

   I am facing problem i= n visualizing G09 output. Gaussview (not reading vibration),Avagadro (not r= eading optimisation steps),chemcraft(not reading vibration and optimization= steps).These all were suitable for G03 output.Please suggest me that what = should I do.Is there any free software for visualizing all results.
=0A<= br> Your suggestions will be  appreciated and helpful for  o= thers too.........









Wit= h regards;
 <= font size=3D"4">ABHISHEK SHAHI
=0A  IISc bangalore-12
  Official E-mail: shahi()ipc.iisc.ernet.in=
  CC:  shahi.abhishek1984()gmail.com
=0A=0A

--0-1210661403-1283059045=:64566-- From owner-chemistry@ccl.net Sun Aug 29 02:18:00 2010 From: "Sayed Mesa elsayed.elmes|*|yahoo.com" To: CCL Subject: CCL: Freq job + Restart with larger number of nodes Message-Id: <-42633-100829020508-12950-wXyCtGP3xji3lkSM5J/xXA,+,server.ccl.net> X-Original-From: "Sayed Mesa" Date: Sun, 29 Aug 2010 02:05:07 -0400 Sent to CCL by: "Sayed Mesa" [elsayed.elmes+*+yahoo.com] Dear CCL community I am running Freq Job for large molecule with one node (eight processors). This job so far took 3 days and is still running. If I stopped the job and increase number of nodes to 2 nodes (16 processors)then restart the Freq job, it will start from the beginning or start from the last point reached in the output? Thanks in advance, Sayed From owner-chemistry@ccl.net Sun Aug 29 09:01:00 2010 From: "Latchmi Singh latchmi.singh*|*gmail.com" To: CCL Subject: CCL:G: need opinion Message-Id: <-42634-100829085906-21305-k3+WSptwuVJkeLT9rOkTpw!^!server.ccl.net> X-Original-From: Latchmi Singh Content-Type: multipart/alternative; boundary=0050450167d536e210048ef5e91e Date: Sun, 29 Aug 2010 08:58:57 -0400 MIME-Version: 1.0 Sent to CCL by: Latchmi Singh [latchmi.singh:+:gmail.com] --0050450167d536e210048ef5e91e Content-Type: text/plain; charset=ISO-8859-1 Hi All, I have a question related to Abishek's. I hope u can help. Whereas he wanted to compare two products from the same reaction, I wish to compare the products of two separate reactions. Actually, I'm not even running the reaction. I've simply computed the total energies of two products and I wish to determine which is the more stable (they are derivatives of a parent product and hence I wish to measure stability wrt the parent molecule). Is it practical to use the computed total energy (E) as a measure of the stability of one against the other (lower energy product being the more stable..)? Or alternately the computed Gibbs energy? Or is it essential to measure stability against reactants (that way also taking stoichiometry and different starting materials into account)? Thanks in advance for your assistance. Regards, Latchmi Singh On Sat, Aug 28, 2010 at 12:06 PM, John McKelvey jmmckel- -gmail.com < owner-chemistry_-_ccl.net> wrote: > > Sent to CCL by: John McKelvey [jmmckel ~~ gmail.com] > I would add one comment to Dave's comment... Catalysis can also be > considered an issue, and this may well be covered by Dave's comment on > "reversibility." If I remember correctly the classical case of the > reaction of N2 + 3H2 --> 2NH3. It is thermodynamically favored, but a > catalyst is required. > > John McKelvey > > On Sat, Aug 28, 2010 at 10:47 AM, David Gallagher > gallagher.da^_^gmail.com wrote: > > Hi Abhishek, > > Assuming the reaction is reversible then the reaction should be > > thermodynamically controlled and product(s) should be those with the > lowest > > free energies. However, you would need to include any counter ions, etc. > in > > the calculation to maintain the stoichiometry (total atoms and charges > must > > be the same). It would also be important to consider the appropriate > > solvent, if relevant, particularly with ions. > > If the prevailing product(s) does not have the lowest free energy, then > the > > reaction may be kinetically controlled, so you would need to characterize > > the transition states to compare the barrier heights. There is a guide > to > > reaction modelling at http://www.cacheresearch.com/presentations.html > > David Gallagher > > CAChe Research > > On Sat, Aug 28, 2010 at 4:00 AM, abhishekdc~!~iitb.ac.in > > wrote: > >> > >> Sent to CCL by: abhishekdc**iitb.ac.in > >> Dear All, > >> Let consider that there is a possibility of formation of two product > >> > from any reaction: A (charge=1, multiplicity=1) and B (charge=2, > >> multiplicity=1) but actually only A is formed. Is there any way to > >> explain above phenomenon computationally ?? Can I compare there energies > >> under equivalent atom condition (Probably not) ?? Can I say something > >> > from there computed delta G value ?? Is there any other way to handle > >> these situation computationally?? > >> > >> I am using Gaussian 03 and many thanks in advance. > >> > >> Sincerely, > >> Abhishek>> E-mail to subscribers: CHEMISTRY!A!ccl.net or use:> >> > >> E-mail to administrators: CHEMISTRY-REQUEST!A!ccl.net or use>> > >> > > > > > > > > -- > John McKelvey > 10819 Middleford Pl > Ft Wayne, IN 46818 > 260-489-2160 > jmmckel*|*gmail.com> > > --0050450167d536e210048ef5e91e Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable Hi All,

I have a question related to Abishek's. I ho= pe u can help.

Whereas he wanted to compare two pr= oducts from the same reaction, I wish to compare the products of two separa= te reactions. Actually, I'm not even running the reaction. I've sim= ply computed the total energies of two products and I wish to determine whi= ch is the more stable (they are derivatives of a parent product and hence I= wish to measure stability wrt the parent molecule).=A0Is it practical to u= se the computed total energy (E) as a measure of the stability of one again= st the other (lower energy product being the more stable..)? Or alternately= the computed Gibbs energy? Or is it essential to measure stability against= reactants (that way also taking stoichiometry and different starting mater= ials into account)?

Thanks in advance for your assistance.

Regards,
Latchmi Singh

On Sat, Aug 28, 2010 at 12:06 PM, John McKelvey jmmckel- -gmail.com <owner-chemistry_-_ccl.net> wrote:<= br>

Sent to CCL by: John McKelvey [jmmckel ~~ gmail.com]
I would add one comment to Dave's comment... Catalysis can also be
considered an issue, and this may well be covered by Dave's comment on<= br> "reversibility." =A0If I remember correctly the classical case of= the
reaction of N2 + 3H2 --> 2NH3. =A0It is thermodynamically favored, but a=
catalyst is required.

John McKelvey

On Sat, Aug 28, 2010 at 10:47 AM, David Gallagher
gallagher.da^_^gmail.com= <owner-chemistry*|*ccl.net= > wrote:
> Hi Abhishek,
> Assuming the reaction is reversible then the reaction should be
> thermodynamically controlled and product(s) should be those with the l= owest
> free energies. =A0However, you would need to include any counter ions,= etc. in
> the calculation to maintain the stoichiometry (total atoms and charges= must
> be the same). =A0It would also be important to consider the appropriat= e
> solvent, if relevant, particularly with ions.
> If the prevailing product(s) does not have the lowest free energy, the= n the
> reaction may be kinetically controlled, so you would need to character= ize
> the transition states to compare the barrier heights. =A0There is a gu= ide to
> reaction modelling at http://www.cacheresearch.com/presentations.ht= ml
> David Gallagher
> CAChe Research
> On Sat, Aug 28, 2010 at 4:00 AM, abhishekdc~!~iitb.ac.in
> <owner-chemistry!A!ccl= .net> wrote:
>>
>> Sent to CCL by: abhishekdc**iitb.ac.in
>> Dear All,
>> =A0Let consider that there is a possibility of formation of two pr= oduct
>> > from any reaction: A (charge=3D1, multiplicity=3D1) and B (ch= arge=3D2,
>> multiplicity=3D1) but actually only A is formed. Is there any way = to
>> explain above phenomenon computationally ?? Can I compare there en= ergies
>> under equivalent atom condition (Probably not) ?? =A0Can I say som= ething
>> > from there computed delta G value ?? Is there any other way t= o handle
>> these situation computationally??
>>
>> I am using Gaussian 03 and many thanks in advance.
>>
>> Sincerely,
>> Abhishek>> E-mail to subscribers: CHEMISTRY!A!ccl.net or use:
>> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/sen= d_ccl_message
>>
>> E-mail to administrators: CHEMISTRY-REQUEST!A!ccl.net or use
>> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/send_ccl_messag= e>> =A0 =A0 =A0http://www.ccl.net/chemistry/sub_unsub.shtml&g= t;> =A0 =A0 =A0http://www.ccl.net/spammers.txt>>
>>
>
>



--
John McKelvey
10819 Middleford Pl
Ft Wayne, IN 46818
260-489-2160
jmmckel*|*gmail.com



-=3D This is automatically added to each message by the mailing script =3D-=
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--0050450167d536e210048ef5e91e-- From owner-chemistry@ccl.net Sun Aug 29 11:12:00 2010 From: "Sergio Emanuel Galembeck segalemb[#]usp.br" To: CCL Subject: CCL: Freq job + Restart with larger number of nodes Message-Id: <-42635-100829082510-16740-vwfZl1HNQJqJYsgQ95bsMA]~[server.ccl.net> X-Original-From: Sergio Emanuel Galembeck Content-Disposition: inline Content-Transfer-Encoding: 7bit Content-Type: text/plain; charset=ISO-8859-1; DelSp="Yes"; format="flowed" Date: Sun, 29 Aug 2010 09:25:01 -0300 MIME-Version: 1.0 Sent to CCL by: Sergio Emanuel Galembeck [segalemb+*+usp.br] Dear Said, Normally frequency jobs are not restartable, except if it is a numerical one. Best regards, Sergio Citando "Sayed Mesa elsayed.elmes|*|yahoo.com" : > > Sent to CCL by: "Sayed Mesa" [elsayed.elmes+*+yahoo.com] > Dear CCL community > > I am running Freq Job for large molecule with one node (eight > processors). This job so far took 3 days and is still running. If I > stopped the job and increase number of nodes to 2 nodes (16 > processors)then restart the Freq job, it will start from the > beginning or start from the last point reached in the output? > > Thanks in advance, > > Sayed> > > From owner-chemistry@ccl.net Sun Aug 29 12:55:00 2010 From: "David Gallagher gallagher.da{}gmail.com" To: CCL Subject: CCL:G: need opinion Message-Id: <-42636-100829120155-4226-dooTuAfihat7FlqG4NoVUg_._server.ccl.net> X-Original-From: David Gallagher Content-Type: multipart/alternative; boundary=0016e6480d1efc02bc048ef876f2 Date: Sun, 29 Aug 2010 09:01:46 -0700 MIME-Version: 1.0 Sent to CCL by: David Gallagher [gallagher.da#%#gmail.com] --0016e6480d1efc02bc048ef876f2 Content-Type: text/plain; charset=ISO-8859-1 Hi Latchmi, Actually, your scenario would seem to be the same as my interpretation of Abhishek's question, i.e. reactant A can undergo two possible competing reactions to product B and product C. Comparing the free energies of B and C (as detailed previously) should be sufficient to estimate the ratio of products if the reactions are reversible, i.e. thermodynamically controlled. However, if the reactions are kinetically controlled, then you need to compare the activation energies for each reaction, i.e. the transition state energies minus the reactant energy A. However, finding the correct transition states is not always trivial (see the guide to transition state modelling about 2/3 down the page at http://www.cacheresearch.com/presentations.html ) . The lowest activation energy should offer the fastest reaction path and hence, lead to the major product. The activation energy plugs into the Arrhenius equation to give the reaction rates, although, this step is not necessary for a qualitative answer. David Gallagher CACheResearch.com On Sun, Aug 29, 2010 at 5:58 AM, Latchmi Singh latchmi.singh*|*gmail.com < owner-chemistry]~[ccl.net> wrote: > Hi All, > > I have a question related to Abishek's. I hope u can help. > > Whereas he wanted to compare two products from the same reaction, I wish to > compare the products of two separate reactions. Actually, I'm not even > running the reaction. I've simply computed the total energies of two > products and I wish to determine which is the more stable (they are > derivatives of a parent product and hence I wish to measure stability wrt > the parent molecule). Is it practical to use the computed total energy (E) > as a measure of the stability of one against the other (lower energy product > being the more stable..)? Or alternately the computed Gibbs energy? Or is it > essential to measure stability against reactants (that way also taking > stoichiometry and different starting materials into account)? > > Thanks in advance for your assistance. > > Regards, > Latchmi Singh > > On Sat, Aug 28, 2010 at 12:06 PM, John McKelvey jmmckel- -gmail.com < > owner-chemistry#,#ccl.net > wrote: > >> >> Sent to CCL by: John McKelvey [jmmckel ~~ gmail.com] >> I would add one comment to Dave's comment... Catalysis can also be >> considered an issue, and this may well be covered by Dave's comment on >> "reversibility." If I remember correctly the classical case of the >> reaction of N2 + 3H2 --> 2NH3. It is thermodynamically favored, but a >> catalyst is required. >> >> John McKelvey >> >> On Sat, Aug 28, 2010 at 10:47 AM, David Gallagher >> gallagher.da^_^gmail.com wrote: >> > Hi Abhishek, >> > Assuming the reaction is reversible then the reaction should be >> > thermodynamically controlled and product(s) should be those with the >> lowest >> > free energies. However, you would need to include any counter ions, >> etc. in >> > the calculation to maintain the stoichiometry (total atoms and charges >> must >> > be the same). It would also be important to consider the appropriate >> > solvent, if relevant, particularly with ions. >> > If the prevailing product(s) does not have the lowest free energy, then >> the >> > reaction may be kinetically controlled, so you would need to >> characterize >> > the transition states to compare the barrier heights. There is a guide >> to >> > reaction modelling at http://www.cacheresearch.com/presentations.html >> > David Gallagher >> > CAChe Research >> > On Sat, Aug 28, 2010 at 4:00 AM, abhishekdc~!~iitb.ac.in >> > wrote: >> >> >> >> Sent to CCL by: abhishekdc**iitb.ac.in >> >> Dear All, >> >> Let consider that there is a possibility of formation of two product >> >> > from any reaction: A (charge=1, multiplicity=1) and B (charge=2, >> >> multiplicity=1) but actually only A is formed. Is there any way to >> >> explain above phenomenon computationally ?? Can I compare there >> energies >> >> under equivalent atom condition (Probably not) ?? Can I say something >> >> > from there computed delta G value ?? Is there any other way to handle >> >> these situation computationally?? >> >> >> >> I am using Gaussian 03 and many thanks in advance. >> >> >> >> Sincerely, >> >> Abhishek>> E-mail to subscribers: CHEMISTRY!A!ccl.net or use:>> >> >> >> E-mail to administrators: CHEMISTRY-REQUEST!A!ccl.net or use>> >> >> >> > >> > >> >> >> >> -- >> John McKelvey >> 10819 Middleford Pl >> Ft Wayne, IN 46818 >> 260-489-2160 >> jmmckel*|*gmail.com>> E-mail to subscribers: CHEMISTRY#,#ccl.net or >> use:>> >> E-mail to administrators: CHEMISTRY-REQUEST#,#ccl.netor use>> >> >> > --0016e6480d1efc02bc048ef876f2 Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable
Hi Latchmi,
Actually, your scenario would seem to be the same as my interpr= etation of Abhishek's question, i.e. reactant A can undergo two possibl= e competing reactions to product B and product C. =A0Comparing the free ene= rgies of B and C (as detailed previously) should be sufficient to estimate = the ratio of products if the reactions are reversible, i.e. thermodynamical= ly controlled. =A0

However, if the reactions are kinetically controlled, t= hen you need to compare the activation energies for each reaction, i.e. the= transition state energies minus the reactant energy A. =A0However, finding= the correct transition states is not always trivial (see the guide to tran= sition state modelling about 2/3 down the page athttp://www.cacheresearch.com/presentations.html=A0) . =A0The= lowest activation energy should offer the fastest reaction path and hence,= lead to the major product. =A0The activation energy plugs into the Arrheni= us equation to give the reaction rates, although, this step is not necessar= y for a qualitative answer.

David Gallagher
CAChe= Research.com

On Sun, Aug = 29, 2010 at 5:58 AM, Latchmi Singh latchmi.singh*|*gmail.com <owner-chemistry]~[ccl.net> wrote:
Hi All,

I have a questio= n related to Abishek's. I hope u can help.

Whe= reas he wanted to compare two products from the same reaction, I wish to co= mpare the products of two separate reactions. Actually, I'm not even ru= nning the reaction. I've simply computed the total energies of two prod= ucts and I wish to determine which is the more stable (they are derivatives= of a parent product and hence I wish to measure stability wrt the parent m= olecule).=A0Is it practical to use the computed total energy (E) as a measu= re of the stability of one against the other (lower energy product being th= e more stable..)? Or alternately the computed Gibbs energy? Or is it essent= ial to measure stability against reactants (that way also taking stoichiome= try and different starting materials into account)?

Thanks in advance for your assistance.

Regards,
Latchmi Singh

On Sat, Aug 28, 2010 at 12:06 PM, John McKelvey jmmckel- -gmail.com = <owner-= chemistry#,#ccl.net> wrote:

Sent to CCL by: John McKelvey [jmmckel ~~ gmail.com]
I would add one comment to Dave's comment... Catalysis can also be
considered an issue, and this may well be covered by Dave's comment on<= br> "reversibility." =A0If I remember correctly the classical case of= the
reaction of N2 + 3H2 --> 2NH3. =A0It is thermodynamically favored, but a=
catalyst is required.

John McKelvey

On Sat, Aug 28, 2010 at 10:47 AM, David Gallagher
gallagher.da^_^gmail.com= <owner-chemistry*|*ccl.net= > wrote:
> Hi Abhishek,
> Assuming the reaction is reversible then the reaction should be
> thermodynamically controlled and product(s) should be those with the l= owest
> free energies. =A0However, you would need to include any counter ions,= etc. in
> the calculation to maintain the stoichiometry (total atoms and charges= must
> be the same). =A0It would also be important to consider the appropriat= e
> solvent, if relevant, particularly with ions.
> If the prevailing product(s) does not have the lowest free energy, the= n the
> reaction may be kinetically controlled, so you would need to character= ize
> the transition states to compare the barrier heights. =A0There is a gu= ide to
> reaction modelling at http://www.cacheresearch.com/presentations.ht= ml
> David Gallagher
> CAChe Research
> On Sat, Aug 28, 2010 at 4:00 AM, abhishekdc~!~iitb.ac.in
> <owner-chemistry!A!ccl= .net> wrote:
>>
>> Sent to CCL by: abhishekdc**iitb.ac.in
>> Dear All,
>> =A0Let consider that there is a possibility of formation of two pr= oduct
>> > from any reaction: A (charge=3D1, multiplicity=3D1) and B (ch= arge=3D2,
>> multiplicity=3D1) but actually only A is formed. Is there any way = to
>> explain above phenomenon computationally ?? Can I compare there en= ergies
>> under equivalent atom condition (Probably not) ?? =A0Can I say som= ething
>> > from there computed delta G value ?? Is there any other way t= o handle
>> these situation computationally??
>>
>> I am using Gaussian 03 and many thanks in advance.
>>
>> Sincerely,
>> Abhishek>> E-mail to subscribers: CHEMISTRY!A!ccl.net or use:
>> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/send_ccl_message=
>>
>> E-mail to administrators: CHEMISTRY-REQUEST!A!ccl.net or use
>> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/send_ccl_messag= e>> =A0 =A0 =A0http://www.ccl.net/chemistry/sub_unsub.shtml&g= t;> =A0 =A0 =A0http://www.ccl.net/spammers.txt>>
>>
>
>



--
John McKelvey
10819 Middleford Pl
Ft Wayne, IN 46818
260-489-2160
jmmckel*|*gmail.com



-=3D This is automatically added to each message by the mailing script =3D-=
To recover the email address of the author of the message, pleas= e change
E-mail to subscribers: CHEMISTRY#,#ccl.net or use:
E-mail to administrators: CHEMISTRY-REQUEST#,#ccl.net or use


--0016e6480d1efc02bc048ef876f2-- From owner-chemistry@ccl.net Sun Aug 29 13:31:01 2010 From: "David Gallagher gallagher.da|,|gmail.com" To: CCL Subject: CCL:G: need opinion Message-Id: <-42637-100829124307-20284-/CNABjWP+3T+vqJhpMnUOQ^server.ccl.net> X-Original-From: David Gallagher Content-Type: multipart/alternative; boundary=000325574d6669b3e0048ef90ad4 Date: Sun, 29 Aug 2010 09:42:59 -0700 MIME-Version: 1.0 Sent to CCL by: David Gallagher [gallagher.da~~gmail.com] --000325574d6669b3e0048ef90ad4 Content-Type: text/plain; charset=ISO-8859-1 Hi Latchmi, Actually, your scenario would seem to be the same as my interpretation of Abhishek's question, i.e. reactant A can undergo two possible competing reactions to product B and product C. Comparing the free energies of B and C (as detailed previously) should be sufficient to estimate the ratio of products if the reactions are reversible, i.e. thermodynamically controlled. However, if the reactions are kinetically controlled, then you need to compare the activation energies for each reaction, i.e. the transition state energies minus the reactant energy A. However, finding the correct transition states is not always trivial (see the guide to transition state modelling about 2/3 down the page at http://www.cacheresearch.com/presentations.html ) . The lowest activation energy should offer the fastest reaction path and hence, lead to the major product. The activation energy plugs into the Arrhenius equation to give the reaction rates, although, this step is not necessary for a qualitative answer. David Gallagher CACheResearch.com On Sun, Aug 29, 2010 at 9:37 AM, David Gallagher wrote: > Hi Latchmi, > > Actually, your scenario would seem to be the same as my interpretation of > Abhishek's question, i.e. reactant A can undergo two possible competing > reactions to product B and product C. Comparing the free energies of B and > C (as detailed previously) should be sufficient to estimate the ratio of > products if the reactions are reversible, i.e. thermodynamically controlled. > > > However, if the reactions are kinetically controlled, then you need to > compare the activation energies for each reaction, i.e. the transition state > energies minus the reactant energy A. However, finding the correct > transition states is not always trivial (see the guide to transition state > modelling about 2/3 down the page at > http://www.cacheresearch.com/presentations.html ) . The lowest activation > energy should offer the fastest reaction path and hence, lead to the major > product. The activation energy plugs into the Arrhenius equation to give > the reaction rates, although, this step is not necessary for a qualitative > answer. > > David Gallagher > CACheResearch.com > > On Sun, Aug 29, 2010 at 5:58 AM, Latchmi Singh latchmi.singh*|*gmail.com < > owner-chemistry#,#ccl.net> wrote: > >> Hi All, >> >> I have a question related to Abishek's. I hope u can help. >> >> Whereas he wanted to compare two products from the same reaction, I wish >> to compare the products of two separate reactions. Actually, I'm not even >> running the reaction. I've simply computed the total energies of two >> products and I wish to determine which is the more stable (they are >> derivatives of a parent product and hence I wish to measure stability wrt >> the parent molecule). Is it practical to use the computed total energy (E) >> as a measure of the stability of one against the other (lower energy product >> being the more stable..)? Or alternately the computed Gibbs energy? Or is it >> essential to measure stability against reactants (that way also taking >> stoichiometry and different starting materials into account)? >> >> Thanks in advance for your assistance. >> >> Regards, >> Latchmi Singh >> >> On Sat, Aug 28, 2010 at 12:06 PM, John McKelvey jmmckel- -gmail.com < >> owner-chemistry#,#ccl.net > wrote: >> >>> >>> Sent to CCL by: John McKelvey [jmmckel ~~ gmail.com] >>> I would add one comment to Dave's comment... Catalysis can also be >>> considered an issue, and this may well be covered by Dave's comment on >>> "reversibility." If I remember correctly the classical case of the >>> reaction of N2 + 3H2 --> 2NH3. It is thermodynamically favored, but a >>> catalyst is required. >>> >>> John McKelvey >>> >>> On Sat, Aug 28, 2010 at 10:47 AM, David Gallagher >>> gallagher.da^_^gmail.com wrote: >>> > Hi Abhishek, >>> > Assuming the reaction is reversible then the reaction should be >>> > thermodynamically controlled and product(s) should be those with the >>> lowest >>> > free energies. However, you would need to include any counter ions, >>> etc. in >>> > the calculation to maintain the stoichiometry (total atoms and charges >>> must >>> > be the same). It would also be important to consider the appropriate >>> > solvent, if relevant, particularly with ions. >>> > If the prevailing product(s) does not have the lowest free energy, then >>> the >>> > reaction may be kinetically controlled, so you would need to >>> characterize >>> > the transition states to compare the barrier heights. There is a guide >>> to >>> > reaction modelling at http://www.cacheresearch.com/presentations.html >>> > David Gallagher >>> > CAChe Research >>> >> > > On Sat, Aug 28, 2010 at 4:00 AM, abhishekdc~!~iitb.ac.in >>> > wrote: >>> >> >>> >> Sent to CCL by: abhishekdc**iitb.ac.in >>> >> Dear All, >>> >> Let consider that there is a possibility of formation of two product >>> >> > from any reaction: A (charge=1, multiplicity=1) and B (charge=2, >>> >> multiplicity=1) but actually only A is formed. Is there any way to >>> >> explain above phenomenon computationally ?? Can I compare there >>> energies >>> >> under equivalent atom condition (Probably not) ?? Can I say something >>> >> > from there computed delta G value ?? Is there any other way to >>> handle >>> >> these situation computationally?? >>> >> >>> >> I am using Gaussian 03 and many thanks in advance. >>> >> >>> >> Sincerely, >>> >> Abhishek> >> >> --000325574d6669b3e0048ef90ad4 Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable
Hi Latchmi,

Actually, your scenar= io would seem to be the same as my interpretation of Abhishek's questio= n, i.e. reactant A can undergo two possible competing reactions to product = B and product C. =A0Comparing the free energies of B and C (as detailed pre= viously) should be sufficient to estimate the ratio of products if the reac= tions are reversible, i.e. thermodynamically controlled. =A0

However, if the reactions are kinetically controlled, then you nee= d to compare the activation energies for each reaction, i.e. the transition= state energies minus the reactant energy A. =A0However, finding the correc= t transition states is not always trivial (see the guide to transition stat= e modelling about 2/3 down the page at=A0http://www.cacheresearch.com/presentations.html<= /a>=A0) . =A0The lowest activation energy should offer the fastest reaction= path and hence, lead to the major product. =A0The activation energy plugs = into the Arrhenius equation to give the reaction rates, although, this step= is not necessary for a qualitative answer.

David Gallagher
CACheResearch.com
On Sun, Aug 29, 2010 at 9:37 AM, David Gallagher <gallagher.da#,#gmail.com> wrote:
Hi Latchmi,

= Actually, your scenario would seem to be the same as my interpretation of A= bhishek's question, i.e. reactant A can undergo two possible competing = reactions to product B and product C. =A0Comparing the free energies of B a= nd C (as detailed previously) should be sufficient to estimate the ratio of= products if the reactions are reversible, i.e. thermodynamically controlle= d. =A0

However, if = the reactions are kinetically controlled, then you need to compare the acti= vation energies for each reaction, i.e. the transition state energies minus= the reactant energy A. =A0However, finding the correct transition states i= s not always trivial (see the guide to transition state modelling about 2/3= down the page at=A0http://www.cacheresearch.com= /presentations.html=A0) . =A0The lowest activation energy should= offer the fastest reaction path and hence, lead to the major product. =A0T= he activation energy plugs into the Arrhenius equation to give the reaction= rates, although, this step is not necessary for a qualitative answer.

David Gallagher
= CACheResearch.com

On Sun, Aug 29, 2010 at 5:58 AM, Latchmi Singh latchmi.si= ngh*|*gmail.com <= owner-chemistry#,#ccl.net> wrote:
Hi All,

I have a question= related to Abishek's. I hope u can help.

Wher= eas he wanted to compare two products from the same reaction, I wish to com= pare the products of two separate reactions. Actually, I'm not even run= ning the reaction. I've simply computed the total energies of two produ= cts and I wish to determine which is the more stable (they are derivatives = of a parent product and hence I wish to measure stability wrt the parent mo= lecule).=A0Is it practical to use the computed total energy (E) as a measur= e of the stability of one against the other (lower energy product being the= more stable..)? Or alternately the computed Gibbs energy? Or is it essenti= al to measure stability against reactants (that way also taking stoichiomet= ry and different starting materials into account)?

Thanks in advance for your assistance.

Regards,
Latchmi Singh

On Sat, Aug 28, 2010 at 12:06 PM, John McKelvey jmmckel- -gmail.com = <owner-= chemistry#,#ccl.net> wrote:

Sent to CCL by: John McKelvey [jmmckel ~~ gmail.com]
I would add one comment to Dave's comment... Catalysis can also be
considered an issue, and this may well be covered by Dave's comment on<= br> "reversibility." =A0If I remember correctly the classical case of= the
reaction of N2 + 3H2 --> 2NH3. =A0It is thermodynamically favored, but a=
catalyst is required.

John McKelvey

On Sat, Aug 28, 2010 at 10:47 AM, David Gallagher
gallagher.da^_^gmail.com= <owner-chemistry*|*ccl.net= > wrote:
> Hi Abhishek,
> Assuming the reaction is reversible then the reaction should be
> thermodynamically controlled and product(s) should be those with the l= owest
> free energies. =A0However, you would need to include any counter ions,= etc. in
> the calculation to maintain the stoichiometry (total atoms and charges= must
> be the same). =A0It would also be important to consider the appropriat= e
> solvent, if relevant, particularly with ions.
> If the prevailing product(s) does not have the lowest free energy, the= n the
> reaction may be kinetically controlled, so you would need to character= ize
> the transition states to compare the barrier heights. =A0There is a gu= ide to
> reaction modelling at http://www.cacheresearch.com/presentations.ht= ml
> David Gallagher
> CAChe Research
=A0
> On Sat, Aug 28, 2010 at 4:00 AM, abhishekdc~!~iitb.ac.in
> <owner-chemistry!A!ccl= .net> wrote:
>>
>> Sent to CCL by: abhishekdc**iitb.ac.in
>> Dear All,
>> =A0Let consider that there is a possibility of formation of two pr= oduct
>> > from any reaction: A (charge=3D1, multiplicity=3D1) and B (ch= arge=3D2,
>> multiplicity=3D1) but actually only A is formed. Is there any way = to
>> explain above phenomenon computationally ?? Can I compare there en= ergies
>> under equivalent atom condition (Probably not) ?? =A0Can I say som= ething
>> > from there computed delta G value ?? Is there any other way t= o handle
>> these situation computationally??
>>
>> I am using Gaussian 03 and many thanks in advance.
>>
>> Sincerely,
>> Abhishek>
--000325574d6669b3e0048ef90ad4-- From owner-chemistry@ccl.net Sun Aug 29 14:06:01 2010 From: "David Gallagher gallagher.da(!)gmail.com" To: CCL Subject: CCL:G: need opinion Message-Id: <-42638-100829112605-24522-MGVKbZWu7TYl2TKg1fYL0w!A!server.ccl.net> X-Original-From: David Gallagher Content-Type: multipart/alternative; boundary=0016364ecc6cf0580b048ef7f600 Date: Sun, 29 Aug 2010 08:25:57 -0700 MIME-Version: 1.0 Sent to CCL by: David Gallagher [gallagher.da]=[gmail.com] --0016364ecc6cf0580b048ef7f600 Content-Type: text/plain; charset=ISO-8859-1 Hi Latchmi, Actually, your scenario would seem to be the same as my interpretation of Abhishek's question, i.e. reactant A can undergo two possible competing reactions to product B and product C. Comparing the free energies of B and C (as detailed previously) should be sufficient to estimate the ratio of products if the reactions are reversible, i.e. thermodynamically controlled. However, if the reactions are kinetically controlled, then you need to compare the activation energies for each reaction, i.e. the transition state energies minus the reactant energy A. However, finding the correct transition states is not always trivial (see the guide to transition state modelling about 2/3 down the page at http://www.cacheresearch.com/presentations.html ) . The lowest activation energy should offer the fastest reaction path and hence, lead to the major product. The activation energy plugs into the Arrhenius equation to give the reaction rates, although, this step is not necessary for a qualitative answer. David Gallagher CACheResearch.com On Sun, Aug 29, 2010 at 5:58 AM, Latchmi Singh latchmi.singh*|*gmail.com < owner-chemistry=-=ccl.net> wrote: > Hi All, > > I have a question related to Abishek's. I hope u can help. > > Whereas he wanted to compare two products from the same reaction, I wish to > compare the products of two separate reactions. Actually, I'm not even > running the reaction. I've simply computed the total energies of two > products and I wish to determine which is the more stable (they are > derivatives of a parent product and hence I wish to measure stability wrt > the parent molecule). Is it practical to use the computed total energy (E) > as a measure of the stability of one against the other (lower energy product > being the more stable..)? Or alternately the computed Gibbs energy? Or is it > essential to measure stability against reactants (that way also taking > stoichiometry and different starting materials into account)? > > Thanks in advance for your assistance. > > Regards, > Latchmi Singh > > On Sat, Aug 28, 2010 at 12:06 PM, John McKelvey jmmckel- -gmail.com < > owner-chemistry#,#ccl.net > wrote: > >> >> Sent to CCL by: John McKelvey [jmmckel ~~ gmail.com] >> I would add one comment to Dave's comment... Catalysis can also be >> considered an issue, and this may well be covered by Dave's comment on >> "reversibility." If I remember correctly the classical case of the >> reaction of N2 + 3H2 --> 2NH3. It is thermodynamically favored, but a >> catalyst is required. >> >> John McKelvey >> >> On Sat, Aug 28, 2010 at 10:47 AM, David Gallagher >> gallagher.da^_^gmail.com wrote: >> > Hi Abhishek, >> > Assuming the reaction is reversible then the reaction should be >> > thermodynamically controlled and product(s) should be those with the >> lowest >> > free energies. However, you would need to include any counter ions, >> etc. in >> > the calculation to maintain the stoichiometry (total atoms and charges >> must >> > be the same). It would also be important to consider the appropriate >> > solvent, if relevant, particularly with ions. >> > If the prevailing product(s) does not have the lowest free energy, then >> the >> > reaction may be kinetically controlled, so you would need to >> characterize >> > the transition states to compare the barrier heights. There is a guide >> to >> > reaction modelling at http://www.cacheresearch.com/presentations.html >> > David Gallagher >> > CAChe Research >> > On Sat, Aug 28, 2010 at 4:00 AM, abhishekdc~!~iitb.ac.in >> > wrote: >> >> >> >> Sent to CCL by: abhishekdc**iitb.ac.in >> >> Dear All, >> >> Let consider that there is a possibility of formation of two product >> >> > from any reaction: A (charge=1, multiplicity=1) and B (charge=2, >> >> multiplicity=1) but actually only A is formed. Is there any way to >> >> explain above phenomenon computationally ?? Can I compare there >> energies >> >> under equivalent atom condition (Probably not) ?? Can I say something >> >> > from there computed delta G value ?? Is there any other way to handle >> >> these situation computationally?? >> >> >> >> I am using Gaussian 03 and many thanks in advance. >> >> >> >> Sincerely, >> >> Abhishek>> E-mail to subscribers: CHEMISTRY!A!ccl.net or use:>> >> >> >> E-mail to administrators: CHEMISTRY-REQUEST!A!ccl.net or use>> >> >> >> > >> > >> >> >> >> -- >> John McKelvey >> 10819 Middleford Pl >> Ft Wayne, IN 46818 >> 260-489-2160 >> jmmckel*|*gmail.com>> E-mail to subscribers: CHEMISTRY#,#ccl.net or >> use:>> >> E-mail to administrators: CHEMISTRY-REQUEST#,#ccl.netor use>> >> >> > --0016364ecc6cf0580b048ef7f600 Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable
Hi Latchmi,

Actually, your scenario would see= m to be the same as my interpretation of Abhishek's question, i.e. reac= tant A can undergo two possible competing reactions to product B and produc= t C. =A0Comparing the free energies of B and C (as detailed previously) sho= uld be sufficient to estimate the ratio of products if the reactions are re= versible, i.e. thermodynamically controlled. =A0

However, if the reactions are kinetically controlled, t= hen you need to compare the activation energies for each reaction, i.e. the= transition state energies minus the reactant energy A. =A0However, finding= the correct transition states is not always trivial (see the guide to tran= sition state modelling about 2/3 down the page at http://www.cacheresearch.com/presentatio= ns.html ) . =A0The lowest activation energy should offer the fastest re= action path and hence, lead to the major product. =A0The activation energy = plugs into the Arrhenius equation to give the reaction rates, although, thi= s step is not necessary for a qualitative answer.

David Gallagher
CACheResearch.com
<= br>

On Sun, Aug 29, 2010 at 5:58 AM, La= tchmi Singh latchmi.singh*|*gmail.com <owner-chemistr= y=-=ccl.net> wrote:
Hi All,

I have a questio= n related to Abishek's. I hope u can help.

Whe= reas he wanted to compare two products from the same reaction, I wish to co= mpare the products of two separate reactions. Actually, I'm not even ru= nning the reaction. I've simply computed the total energies of two prod= ucts and I wish to determine which is the more stable (they are derivatives= of a parent product and hence I wish to measure stability wrt the parent m= olecule).=A0Is it practical to use the computed total energy (E) as a measu= re of the stability of one against the other (lower energy product being th= e more stable..)? Or alternately the computed Gibbs energy? Or is it essent= ial to measure stability against reactants (that way also taking stoichiome= try and different starting materials into account)?

Thanks in advance for your assistance.

Regards,
Latchmi Singh

On Sat, Aug 28, 2010 at 12:06 PM, John McKelvey jmmckel- -gmail.com = <owner-= chemistry#,#ccl.net> wrote:

Sent to CCL by: John McKelvey [jmmckel ~~ gmail.com]
I would add one comment to Dave's comment... Catalysis can also be
considered an issue, and this may well be covered by Dave's comment on<= br> "reversibility." =A0If I remember correctly the classical case of= the
reaction of N2 + 3H2 --> 2NH3. =A0It is thermodynamically favored, but a=
catalyst is required.

John McKelvey

On Sat, Aug 28, 2010 at 10:47 AM, David Gallagher
gallagher.da^_^gmail.com= <owner-chemistry*|*ccl.net= > wrote:
> Hi Abhishek,
> Assuming the reaction is reversible then the reaction should be
> thermodynamically controlled and product(s) should be those with the l= owest
> free energies. =A0However, you would need to include any counter ions,= etc. in
> the calculation to maintain the stoichiometry (total atoms and charges= must
> be the same). =A0It would also be important to consider the appropriat= e
> solvent, if relevant, particularly with ions.
> If the prevailing product(s) does not have the lowest free energy, the= n the
> reaction may be kinetically controlled, so you would need to character= ize
> the transition states to compare the barrier heights. =A0There is a gu= ide to
> reaction modelling at http://www.cacheresearch.com/presentations.ht= ml
> David Gallagher
> CAChe Research
> On Sat, Aug 28, 2010 at 4:00 AM, abhishekdc~!~iitb.ac.in
> <owner-chemistry!A!ccl= .net> wrote:
>>
>> Sent to CCL by: abhishekdc**iitb.ac.in
>> Dear All,
>> =A0Let consider that there is a possibility of formation of two pr= oduct
>> > from any reaction: A (charge=3D1, multiplicity=3D1) and B (ch= arge=3D2,
>> multiplicity=3D1) but actually only A is formed. Is there any way = to
>> explain above phenomenon computationally ?? Can I compare there en= ergies
>> under equivalent atom condition (Probably not) ?? =A0Can I say som= ething
>> > from there computed delta G value ?? Is there any other way t= o handle
>> these situation computationally??
>>
>> I am using Gaussian 03 and many thanks in advance.
>>
>> Sincerely,
>> Abhishek>> E-mail to subscribers: CHEMISTRY!A!ccl.net or use:
>> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/send_ccl_message=
>>
>> E-mail to administrators: CHEMISTRY-REQUEST!A!ccl.net or use
>> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/send_ccl_messag= e>> =A0 =A0 =A0http://www.ccl.net/chemistry/sub_unsub.shtml&g= t;> =A0 =A0 =A0http://www.ccl.net/spammers.txt>>
>>
>
>



--
John McKelvey
10819 Middleford Pl
Ft Wayne, IN 46818
260-489-2160
jmmckel*|*gmail.com



-=3D This is automatically added to each message by the mailing script =3D-=
To recover the email address of the author of the message, pleas= e change
E-mail to subscribers: CHEMISTRY#,#ccl.net or use:
E-mail to administrators: CHEMISTRY-REQUEST#,#ccl.net or use


--0016364ecc6cf0580b048ef7f600-- From owner-chemistry@ccl.net Sun Aug 29 14:41:01 2010 From: "David Gallagher gallagher.da]-[gmail.com" To: CCL Subject: CCL:G: need opinion Message-Id: <-42639-100829123814-31368-1Qo/XhKoDVgHGW1RYaGhBg-$-server.ccl.net> X-Original-From: David Gallagher Content-Type: multipart/alternative; boundary=00032557a0d67cdbf5048ef8f5ce Date: Sun, 29 Aug 2010 09:37:08 -0700 MIME-Version: 1.0 Sent to CCL by: David Gallagher [gallagher.da!^!gmail.com] --00032557a0d67cdbf5048ef8f5ce Content-Type: text/plain; charset=ISO-8859-1 Hi Latchmi, Actually, your scenario would seem to be the same as my interpretation of Abhishek's question, i.e. reactant A can undergo two possible competing reactions to product B and product C. Comparing the free energies of B and C (as detailed previously) should be sufficient to estimate the ratio of products if the reactions are reversible, i.e. thermodynamically controlled. However, if the reactions are kinetically controlled, then you need to compare the activation energies for each reaction, i.e. the transition state energies minus the reactant energy A. However, finding the correct transition states is not always trivial (see the guide to transition state modelling about 2/3 down the page at http://www.cacheresearch.com/presentations.html ) . The lowest activation energy should offer the fastest reaction path and hence, lead to the major product. The activation energy plugs into the Arrhenius equation to give the reaction rates, although, this step is not necessary for a qualitative answer. David Gallagher CACheResearch.com On Sun, Aug 29, 2010 at 5:58 AM, Latchmi Singh latchmi.singh*|*gmail.com < owner-chemistry[-]ccl.net> wrote: > Hi All, > > I have a question related to Abishek's. I hope u can help. > > Whereas he wanted to compare two products from the same reaction, I wish to > compare the products of two separate reactions. Actually, I'm not even > running the reaction. I've simply computed the total energies of two > products and I wish to determine which is the more stable (they are > derivatives of a parent product and hence I wish to measure stability wrt > the parent molecule). Is it practical to use the computed total energy (E) > as a measure of the stability of one against the other (lower energy product > being the more stable..)? Or alternately the computed Gibbs energy? Or is it > essential to measure stability against reactants (that way also taking > stoichiometry and different starting materials into account)? > > Thanks in advance for your assistance. > > Regards, > Latchmi Singh > > On Sat, Aug 28, 2010 at 12:06 PM, John McKelvey jmmckel- -gmail.com < > owner-chemistry#,#ccl.net > wrote: > >> >> Sent to CCL by: John McKelvey [jmmckel ~~ gmail.com] >> I would add one comment to Dave's comment... Catalysis can also be >> considered an issue, and this may well be covered by Dave's comment on >> "reversibility." If I remember correctly the classical case of the >> reaction of N2 + 3H2 --> 2NH3. It is thermodynamically favored, but a >> catalyst is required. >> >> John McKelvey >> >> On Sat, Aug 28, 2010 at 10:47 AM, David Gallagher >> gallagher.da^_^gmail.com wrote: >> > Hi Abhishek, >> > Assuming the reaction is reversible then the reaction should be >> > thermodynamically controlled and product(s) should be those with the >> lowest >> > free energies. However, you would need to include any counter ions, >> etc. in >> > the calculation to maintain the stoichiometry (total atoms and charges >> must >> > be the same). It would also be important to consider the appropriate >> > solvent, if relevant, particularly with ions. >> > If the prevailing product(s) does not have the lowest free energy, then >> the >> > reaction may be kinetically controlled, so you would need to >> characterize >> > the transition states to compare the barrier heights. There is a guide >> to >> > reaction modelling at http://www.cacheresearch.com/presentations.html >> > David Gallagher >> > CAChe Research >> > On Sat, Aug 28, 2010 at 4:00 AM, abhishekdc~!~iitb.ac.in >> > wrote: >> >> >> >> Sent to CCL by: abhishekdc**iitb.ac.in >> >> Dear All, >> >> Let consider that there is a possibility of formation of two product >> >> > from any reaction: A (charge=1, multiplicity=1) and B (charge=2, >> >> multiplicity=1) but actually only A is formed. Is there any way to >> >> explain above phenomenon computationally ?? Can I compare there >> energies >> >> under equivalent atom condition (Probably not) ?? Can I say something >> >> > from there computed delta G value ?? Is there any other way to handle >> >> these situation computationally?? >> >> >> >> I am using Gaussian 03 and many thanks in advance. >> >> >> >> Sincerely, >> >> Abhishek>> E-mail to subscribers: CHEMISTRY!A!ccl.net or use:>> >> >> >> E-mail to administrators: CHEMISTRY-REQUEST!A!ccl.net or use>> >> >> >> > >> > >> >> >> >> -- >> John McKelvey >> 10819 Middleford Pl >> Ft Wayne, IN 46818 >> 260-489-2160 >> jmmckel*|*gmail.com>> E-mail to subscribers: CHEMISTRY#,#ccl.net or >> use:>> >> E-mail to administrators: CHEMISTRY-REQUEST#,#ccl.netor use>> >> >> > --00032557a0d67cdbf5048ef8f5ce Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable
Hi Latchmi,

Actually, your scenar= io would seem to be the same as my interpretation of Abhishek's questio= n, i.e. reactant A can undergo two possible competing reactions to product = B and product C. =A0Comparing the free energies of B and C (as detailed pre= viously) should be sufficient to estimate the ratio of products if the reac= tions are reversible, i.e. thermodynamically controlled. =A0

However, if the reactions are kinetically controlled, then you nee= d to compare the activation energies for each reaction, i.e. the transition= state energies minus the reactant energy A. =A0However, finding the correc= t transition states is not always trivial (see the guide to transition stat= e modelling about 2/3 down the page at=A0http://www.cacheresearch.com/presentations.html<= /a>=A0) . =A0The lowest activation energy should offer the fastest reaction= path and hence, lead to the major product. =A0The activation energy plugs = into the Arrhenius equation to give the reaction rates, although, this step= is not necessary for a qualitative answer.

David Gallagher
CACheResearch.com
On Sun, Aug 29, 2010 at= 5:58 AM, Latchmi Singh latchmi.singh*|*gmail.= com <ow= ner-chemistry[-]ccl.net> wrote:
Hi All,

I have a questio= n related to Abishek's. I hope u can help.

Whe= reas he wanted to compare two products from the same reaction, I wish to co= mpare the products of two separate reactions. Actually, I'm not even ru= nning the reaction. I've simply computed the total energies of two prod= ucts and I wish to determine which is the more stable (they are derivatives= of a parent product and hence I wish to measure stability wrt the parent m= olecule).=A0Is it practical to use the computed total energy (E) as a measu= re of the stability of one against the other (lower energy product being th= e more stable..)? Or alternately the computed Gibbs energy? Or is it essent= ial to measure stability against reactants (that way also taking stoichiome= try and different starting materials into account)?

Thanks in advance for your assistance.

Regards,
Latchmi Singh

On Sat, Aug 28, 2010 at 12:06 PM, John McKelvey jmmckel- -gmail.com = <owner-= chemistry#,#ccl.net> wrote:

Sent to CCL by: John McKelvey [jmmckel ~~ gmail.com]
I would add one comment to Dave's comment... Catalysis can also be
considered an issue, and this may well be covered by Dave's comment on<= br> "reversibility." =A0If I remember correctly the classical case of= the
reaction of N2 + 3H2 --> 2NH3. =A0It is thermodynamically favored, but a=
catalyst is required.

John McKelvey

On Sat, Aug 28, 2010 at 10:47 AM, David Gallagher
gallagher.da^_^gmail.com= <owner-chemistry*|*ccl.net= > wrote:
> Hi Abhishek,
> Assuming the reaction is reversible then the reaction should be
> thermodynamically controlled and product(s) should be those with the l= owest
> free energies. =A0However, you would need to include any counter ions,= etc. in
> the calculation to maintain the stoichiometry (total atoms and charges= must
> be the same). =A0It would also be important to consider the appropriat= e
> solvent, if relevant, particularly with ions.
> If the prevailing product(s) does not have the lowest free energy, the= n the
> reaction may be kinetically controlled, so you would need to character= ize
> the transition states to compare the barrier heights. =A0There is a gu= ide to
> reaction modelling at http://www.cacheresearch.com/presentations.ht= ml
> David Gallagher
> CAChe Research
> On Sat, Aug 28, 2010 at 4:00 AM, abhishekdc~!~iitb.ac.in
> <owner-chemistry!A!ccl= .net> wrote:
>>
>> Sent to CCL by: abhishekdc**iitb.ac.in
>> Dear All,
>> =A0Let consider that there is a possibility of formation of two pr= oduct
>> > from any reaction: A (charge=3D1, multiplicity=3D1) and B (ch= arge=3D2,
>> multiplicity=3D1) but actually only A is formed. Is there any way = to
>> explain above phenomenon computationally ?? Can I compare there en= ergies
>> under equivalent atom condition (Probably not) ?? =A0Can I say som= ething
>> > from there computed delta G value ?? Is there any other way t= o handle
>> these situation computationally??
>>
>> I am using Gaussian 03 and many thanks in advance.
>>
>> Sincerely,
>> Abhishek>> E-mail to subscribers: CHEMISTRY!A!ccl.net or use:
>> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/send_ccl_message=
>>
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>>
>
>



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