# Why do we *subtract* enthalpy of lattice formation?

1. Feb 3, 2014

### aleksbooker

Hello all,

I'm in gen chem 2 and we're going over how to calculate the enthalpy of lattice formation. The way given is to use the Born-Haber process and add the enthalpies of all the steps in between.

e.g. $Na_{(s)} --> Na^+_{(g)} + e^-$ (388kJ)

There are three or four of these, and we combine (add) them to find the enthalpy of reaction. I know how to *do* the problem, subtracting for both sides to determine the value of the missing variable. Here's where I'm confused:

Why is it that everything else is added, and only enthalpy of lattice formation is subtracted?

(enthalpy of sublimation of sodium) + (enthalpy of sodium ionization) + (enthalpy of fluorine atom formation?) + (enthalpy of fluorine ion formation) - (enthalpy of lattice formation) = enthalpy of sodium fluoride reaction

That fourth operation, why is it subtraction and not addition?

2. Feb 4, 2014

### Staff: Mentor

Technically NaF already exists before the solid is formed.

3. Feb 4, 2014

### aleksbooker

@Borek, you mean we subtract NaF because it already exists?

I don't understand what that means. Are we removing it from the sequence?

As far as I understand, it proceeds logically: solid sodium becomes gaseous sodium becomes ionized sodium gas, while diatomic fluorine gas becomes monatomic fluorine gas becomes ionized fluorine gas, and then ionized sodium gas combines with ionized fluorine gas to form a lattice (which requires enthalpy of lattice energy) which becomes sodium fluoride gas (NaF). How does NaF exist before going through this process?