# Quick Enthalpy question

1. Aug 13, 2004

### mrjeffy321

In a reaction, if I have a product/reactant that is aqueous, do I use the values of enthalpy for their ions?

for example:
2Fe (s) + 3NaOCl(aq) --> Fe2O3 (s) + 3NaCl(aq)
which could also be:
2Fe (s) + 3Na-1(aq) + 3O+2(aq) + 3Cl-1(aq) --> Fe2O3 (s) + 3Na+1(aq) + 3Cl-1(aq)

since I cant find a value for the enthalpy of NaOCl, then this would work out nicely to use the ions, and I dont see an aqueous enthalpy value for NaCl, so this is why I am thinking that I should use the ion's enthalpy, am I right.

Last edited: Aug 13, 2004
2. Aug 13, 2004

### mrjeffy321

but then again, if this is true, then I would need to have the values of enthalpy for a sodium ion that is negative 1, rather than positive, and what an oxygen ion?

Last edited: Aug 13, 2004
3. Aug 14, 2004

### GCT

I don't believe so. The enthalpy of formation of the following aqueous compounds is the net enthalpy of formation from the corresponding elements from their most stable states; there are substeps involved, one of these substeps I believe pertains to solvation...formation of ions, then solvation.

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http://groups.msn.com/GeneralChemistryHomework/

4. Aug 14, 2004

### Gokul43201

Staff Emeritus
mrjeffy,

I see you're still rusting, with bleach this time - for your thermit weld, I guess ?

Okay, you've gotten the oxidation states in NaOCl wrong - it should be Na(+1)O(-2)Cl(+1). It's the Chlorine ion that has a weird state. I'm sure the the reduction potential for :
Cl(+1) + 2e- ---> Cl (-1) is documented.

From this, you can calculate the Free Energy change, and from that, the enthalpy change.

But it's not so simple, because you will also have to use the solvation energy for O(-2) as well as the lattice energy for Fe2O3 OR the ionization and lattice energies for Fe - depending on what data you use.

5. Aug 14, 2004

### mrjeffy321

yes, im still rusting, but mostly just trying to organize all this data I have and get rid of all the scrap paper I have laying around, so I can have all the info I could possibly ever want on these reactions without the clutter.

ok, all see what I can do with this method of doing it.