Subject: hydration enthalpy of H+

From: "Mr HG Kruger" <[email-masked]>
Date: Wed, 4 Mar 1998 09:10:21 GMT+0200

Dear colleaques,

I am trying to calculate the hydration enthalpy of H+ using a 
periodic box.  (semiempirical PM3).  The value according to 
literature should be -260.5 kcal/mol.

Without solvent box the total net effect for H.....H2O+  is -141.1 kcal/mol
[ H+ (353.6)  +   H2O (-53.5)  =   H......H2O+   (159.0) ]

Total effect = 159.0 - (353.6 + 53.5)
                    = -141.1 kcal/mol

With solvent box including 15 H2O molecules and H3O+ one gets:
H+  (353.6)  +  16 H2O (16 * -53.5)  =   H.....H2O+  +  15 H2O (-733.5)

Total effect = -733.5 - (353.6 + (16*-53.5))
                     = -231.1 kcal/mol  

(which seems quite close to the experimental value of -260 kcal/mol 
for the hydration enthalpy of H+)

If one use a different approach/equation:
H3O+ (159) + 15H2O (15 * -53.5) = H3O+ solvated in 15H2O (-733.5)

Total effect  =  -733.5 - (159 + (15* -53.5))
                      = -90 kcal/mol

I then used H+ alone in a solvent box of 31 water molecules:
H+ (353.6)  +  31H2O (31* -53.5)  =  H+ solvated (-1693.5)
 Total effect  =  -1693.5 - (353.6 + (31 * -53.5))
                       = -388.6

(Note that if one optimise the solvated H+ one of the water molecules 
would move close to the H+ ion (H-O distance 0.98 Amstrong) ie H3O+ 
forms.)

I would be glad to get advice on the best approach to calculate the 
hydration enthalpy of H+ using a solvent box.

Thanks for your time.
Gert Kruger
__________________________________________________________
Dr HG Kruger, Dept Chemistry, University of Natal,
PO Box 18091, Dalbridge 4014, Durban, South Afica
Tel  +27-31-2602181   Fax  +27-31-2603091
Email  [email-masked] 
__________________________________________________________