Chemical thermodynamics, heat of formation and heat of atomization

[neolayman]

Homework Statement

Sorry if this kind of question is atypical for this forum. I'm trying to understand the physical concepts involved in my chemistry class.

My homework is covering lattice enthalpy, and to get the lattice enthalpy of an ionic compound using the Born-Haber Cycle:

Homework Equations

lattice enthalpy = -(enthalpy of formation of ionic compound) + heat of atomization of the elemental components + enthalpy of ionization of the cation - electron affinity of the anion

The Attempt at a Solution

In the examples that the book gives, it uses the heat of formation values to replace the heat of atomization values. I vaguely understand what heat of formation is, but not in a way that would suggest that it should be equal to the value of heat of atomization.

Is:
heat of atomization = - heat of formation
but for some double negative that exists beneath the lattice equation that I don't see? Is my textbook wrong?


Could anyone help to explain heat of formation as it relates to heat of atomization? I'm not really sure if the author of my text or my professor understands why they are equal in this equation. "Standard state" also seems to constantly come up in the text while the examples all suggest that we should ignore the difference in effects from all other "states" as it were. As a physics major that makes me very uncomfortable.

 

[neolayman]

I've continued on in the chapter, but I still feel as though I'm learning like a parrot and not getting the concepts clearly enough. If my first post here was too confusing to understand, please let me know. The simple form of my question is as follows:

Why does enthalpy of atomization = enthalpy of formation?

 

[Borek]

As far as I can tell - it doesn't. But it is hard to say what they mean without analyzing whole text.

Perhaps try to read about BH cycle in other places (plenty of lecture notes on the web), once you will feel comfortable with the idea it may click.