# [gmx-users] Re: Re: how to calculate the vaporation enthalpy of pure small organic compound (David van der Spoel)

David van der Spoel spoel at xray.bmc.uu.se
Sun Oct 25 13:15:58 CET 2009

```Jinyao Wang wrote:
> Hi David,
> Thank you for your help. I am very sorry to trouble you.
> I have know your means that the gas-phase energy should be considered.
> But I am confused that you said "the gas-phase energy (with com removed) and add kT".
> What is "the gas-phase energy (with com removed) " and Could you talk about the details?
>
Better look it up in a textbook, e.g.Allen & TIldesley. The equations
are also in my paper
J. Phys. Chem. B. 105 pp. 2618-2626 (2001).

>>> Jinyao Wang wrote:
>>>>  Hi gmx-users,
>>>> I want to  calcualted the the vaporation enthalpy of benzaldehyde  I
>>>> think that the the vaporation enthalpy is the sum of non-bond
>>>> interacion among the benzaldehyde molecule.    So I made a NPT system
>>>> simulation including 512 benzaldehyde molecule.  The following is the
>>>> intermolecule nonbond interaction using the g_energy.
>>>>
>>>>      LJ-(SR)= -21128.7 Kj/mol
>>>>      LJ-(LR)= -889.855 Kj/mol
>>>>      Coulomb-(SR)= -3884.6 Kj/mol
>>>>      Coul.-recip = -2261.51 Kj/mol
>>>>   the sum of non-bond interaction = LJ-(SR) + LJ-(LR) + Coulomb-(SR) +
>>>> Coul.-recip = 28164.665 Kj/mol
>>>>    But the experimental value of he vaporation enthalpy = 50.3 Kj/mol
>>>> I have no ideal that why they have so much different. Now I don't know
>>>> how to solve it.
>>>> Any suggestion will be appreciated.
>>>>
>>> These are totals for the intermolecular energy of the system.  I'm
>>> guessing you didn't use "-nmol 512" to calculate these results?
>> And then subtract the gas-phase energy (with com removed) and add kT.
>
>
> 　　　　　　　　Jinyao Wang
> 　　　　　　　　wangjy at ciac.jl.cn
> 　　　　　　　　　　2009-10-24
>
>
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