[UG]

I've been searching the net for a good Born-Haber cycle/diagram and they all seem to be giving me different ways of getting to the final answer. Can someone please check if my working is right for this question:

Calculate the lattice enthalpy of aluminium oxide given that it's enthalpy of formation is -1669 kJ mol^-1

2Al(s) + 1.5O₂(g)  Al₂O₃        Δ_fH= -1669 kJ mol^-1

Here is what I did:

Δ_atH(Al,s)=314 kJ mol^-1
Multiply this by two as there are two of them, this gives 628 kJ mol^-1.
Now we have 2Al(g)

The first, second and third ionisation energies of aluminium are: 577, 1820 and 2740 kJ mol^-1 respectively.
I did: 2(577+1820+2740) = 10274 kJ mol^-1.

The Δ_atH(O₂,g) is 248 kJ mol^-1
I multiplied this by 1.5 to get 372 kJ mol^-1
We now have 3O(g)

The first and second electron affinity of oxygen is -142 and 844 kJ mol^-1 respectively,
I did 3(-142+844) to get 2106 kJ mol^-1

Is this correct? Have I missed anything out?
If it is correct then is the lattice enthalpy: -1669 + 628 +10274 +372 + 2106 = +11711 kJ mol^-1
Thanks 

[Astrokel]

It is good to write out each enthalpies chemical equation before doing its math. Check electron affinity.

[UG]

Okay, so there is:

2Al(s)  2Al(g)   Δ_atH= +314 kJ mol^-1
1.5O₂(g)  3O(g)  Δ_atH= +248 kJ mol^-1

I multiply the first by 2 and the second by 1.5, to get 628 and 372 kJ mol^-1 respectively.

Then there is:

2Al(g)  2Al⁺(g)   Δ_IE1H= 577kJ mol^-1
2Al⁺(g)  2Al²⁺(g)   Δ_IE2H= 1820kJ mol^-1
2Al³⁺(g)  2Al³⁺(g)   Δ_IE3H= 2740kJ mol^-1

2(577+1820+2740) gives 10274 kJ mol^-1

You ask me to check the electron affinity?

http://www.chemguide.co.uk/atoms/properties/eas.html

At the very bottom the values are exactly the same as mine.

Then:

3O(g)  3O^-(g)    Δ_EA1H= -142 kJ mol^-1
3O^-  3O^2-     Δ_EA2H= +844 kJ mol^-1

I did 3 x -142 +3 x 844 = +2106 kJ mol^-1

Adding these up: -1669 + 628 +372 + 10274 + 2106 still gives 11711 kJ mol^-1
So is this value correct?  

[Astrokel]

Ah, sorry, I meant atomization.

The ΔatH(O2,g) is 248 kJ mol-1
I multiplied this by 1.5 to get 372 kJ mol-1
We now have 3O(g)

1/2 O₂ (g)  O (g)   :delta: H_at = +248

[UG]

Thanks,
Does this mean that I should multiply this equation:

1.5O₂(g) 3O(g)  Δ_atH= +248 kJ mol^-1

by 3 instead of 1.5? In which case I will get 248 x 3 = 744 kJ mol^-1
and then the final answer should be, -1669 + 628 +744 + 10274 + 2106 = 12083 kJ mol^-1
Correct? 

[Astrokel]

Yes you got all the values right except that if you were to draw a born-haber cycle, you will realize that it should be,

+1669 + 628 +744 + 10274 + 2106 = LE

[UG]

Thank you, thank you very much 

Another thing I am a bit confused about, in one of my textbooks (C.E.Housecroft & E.C.Constable. Chemistry: an integrated approach p.288) it states that the lattice enthalpy of potassium chloride can be calculated as:

Δ_latticeH^o(KCl,s) = Δ_fH^o(KCl,s) - Δ_fH^o(K⁺,g) - Δ_fH^o(Cl^-,g)   

So if I applied this to my calculations, then should it not be:

Δ_latticeH^o(Al₂O₃,s) = Δ_fH^o(Al₂O₃,s) - Δ_fH^o(Al³⁺,g) - Δ_fH^o(O^2-,g)

In which case I get Δ_fH^o(Al³⁺,g) as=

2Al(s)  2Al(g)   Δ_atH= +314 kJ mol^-1
2Al(g)  2Al⁺(g)   Δ_IE1H= 577 kJ mol^-1
2Al⁺(g)  2Al²⁺(g)   Δ_IE2H= 1820 kJ mol^-1
2Al³⁺(g)  2Al³⁺(g)   Δ_IE3H= 2740 kJ mol^-1   

314 x 2 + 2(577 + 1820 + 2740) = 10902 kJ mol^-1

And the Δ_fH^o(O^2-,g) as=

1.5O₂(g)  3O(g)  Δ_atH= +248 kJ mol^-1
3O(g)  3O^-(g)    Δ_EA1H= -142 kJ mol^-1
3O^-  3O^2-     Δ_EA2H= +844 kJ mol^-1

3 x 248 + 3(-142+844) = 2850 kJ mol^-1

Using these numbers and the enthalpy of formation being -1669 kJ mol^-1 and using the equation:

Δ_latticeH^o(Al₂O₃,s) = Δ_fH^o(Al₂O₃,s) - Δ_fH^o(Al³⁺,g) - Δ_fH^o(O^2-,g)

I end up with:
-1669 - 10902 - 2850 = -15421 kJ mol^-1 

WHY DO I GET A NEGATIVE ANSWER?

[Astrokel]

I believe the confusion is that actually there are two types of lattice energy - Lattice Forming and Lattice breaking. When i took A level, i was taught that lattice forming, A^x+ + B^y+  A_yB_x would be a negative value. But to say it is not entirely true because the problem lies in how the text define lattice energy(whether is lattice forming or breaking) and thus the sign would differ. Have to check up on the current IUPAC definition of lattice energy. Is your text an old or new one?

[UG]

So please correct me if I am wrong  , the lattice enthalpy of formation (energy change when one mole of a solid ionic compound is formed from it's constituent ions in the gas phase) will be negative? Or not always?
And the lattice enthalpy of dissociation (when the ionic compound is split into its ions (are the ions also in the gas phase?  )) will always be positive because bonds are being broken?

----
Oh yeah, my text? 1997 
Jeez, I feel old now 

[Astrokel]

You are correct, it will always be negative and positive respectively. In some text, they define lattice energy as what you have defined in lattice energy of dissociation resulting in confusion over the sign.

http://viswiki.com/en/Lattice_energy

[UG]

Cool, thank you very much Astrokel. 

[UG]

Just getting back to the enthalpy of atomisation,
For oxygen Astrokel wrote it as:

1/2O₂(g) O(g)   

Does that mean for iodine it is written as:

1/2I₂(g) I(g)

OR

I₂(g) 2I(g)

Or neither? 

And for water, is it written:

H₂O(g)  2H(g) + O(g)

[Astrokel]

1/2I2(g) I(g)

This is correct. Enthalpy change of atomization is the heat absorbed when 1 mole of atoms is formed from its element. You are correct about the water too.

[UG]

Enthalpy change of atomization is the heat absorbed when 1 mole of atoms is formed from its element.

ALWAYS EVERYTHING in the gas phase right?