From owner-chemistry@ccl.net Fri Apr  5 01:50:01 2013
From: "John Keller jwkeller|alaska.edu" <owner-chemistry:+:server.ccl.net>
To: CCL
Subject: CCL:G: how to calculate energy in kcal/mol for a particular temperature
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X-Original-From: John Keller <jwkeller^alaska.edu>
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Date: Thu, 4 Apr 2013 21:13:42 -0800
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Sent to CCL by: John Keller [jwkeller[a]alaska.edu]
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Hi Ashutosh,
You can use the Eyring-Polanyi equation to calculate k the unimolecular
rate constant, given Gibbs Free Energy of Activation delGdoubledagger.
Google "Eyring equation wiki" to see the equation written out.

To a good approximation at moderate temperatures, the free energy of
activation is not dependent on temperature. Reactions go faster at higher
temperatures because a higher proportion of molecules have attained a
kinetic energy greater or equal to the free energy of activation.

To estimate free energy of activation, you would need to optimize and do
frequency calculation for the starting compound and transition state, and
take the difference in free energies between these two. Locating the T.S.
is sometimes difficult: if you are using Gaussian, the Opt=TS, or OPT=QST2,
or Opt=QST3 methods are used.

Good luck!
John Keller



On Thu, Apr 4, 2013 at 7:17 PM, ashutosh gupta ashu1809 a gmail.com <
owner-chemistry%x%ccl.net> wrote:

> Dear All,
>
> Normally it is assumed that if the activation barrier is less than 20
> kcal/mol, then the reaction occurs at ROOM TEMPERATURE.
>
> how do we calculate it?
> what is its formula?
>
> how  can  we know corresponding energy in kcal/mol for some other
> temperature say 70 degree Celsius.
>
> thanking you.
>
> kind regards and best wishes
> ashutosh
> varanasi India
>

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<div dir=3D"ltr"><div><div><div><div><div><div>Hi Ashutosh,<br></div>You ca=
n use the Eyring-Polanyi equation to calculate k the unimolecular rate cons=
tant, given Gibbs Free Energy of Activation delGdoubledagger. Google &quot;=
Eyring equation wiki&quot; to see the equation written out.<br>
<br></div>To a good approximation at moderate temperatures, the free energy=
 of activation is not dependent on temperature. Reactions go faster at high=
er temperatures because a higher proportion of molecules have attained a ki=
netic energy greater or equal to the free energy of activation.<br>
<br></div>To estimate free energy of activation, you would need to optimize=
 and do frequency calculation for the starting compound and transition stat=
e, and take the difference in free energies between these two. Locating the=
 T.S. is sometimes difficult: if you are using Gaussian, the Opt=3DTS, or O=
PT=3DQST2, or Opt=3DQST3 methods are used.<br>
<br></div>Good luck!<br></div>John Keller<br></div><br></div><div class=3D"=
gmail_extra"><br><br><div class=3D"gmail_quote">On Thu, Apr 4, 2013 at 7:17=
 PM, ashutosh gupta ashu1809 a <a href=3D"http://gmail.com">gmail.com</a> <=
span dir=3D"ltr">&lt;<a href=3D"mailto:owner-chemistry%x%ccl.net" target=3D"_=
blank">owner-chemistry%x%ccl.net</a>&gt;</span> wrote:<br>
<blockquote class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1p=
x #ccc solid;padding-left:1ex">Dear All,
<div><br></div><div>Normally it is assumed that if the activation barrier i=
s less than 20 kcal/mol, then the reaction occurs at ROOM TEMPERATURE.</div=
><div><br></div><div>how do we calculate it?</div><div>what is its formula?=
</div>

<div><br></div><div>how =A0can =A0we know corresponding energy in kcal/mol =
for some other temperature say 70 degree Celsius.=A0</div><div><br></div><d=
iv>thanking you.</div><div><br></div><div>kind regards and best wishes=A0</=
div>

<div>ashutosh</div><div>varanasi India</div>
</blockquote></div><br></div>

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