[Superboy92]

Calculate the lattice energy for CaH₂, in kJ/mole, using the following information:
EA for H = -73 kJ/mol; BE for hydrogen = 432 kJ/mol
IE₁ for Ca = 591 kJ/mol; IE₂ for Ca = 1146 kJ/mol
ΔH_f for CaH₂ = -186 kJ/mol


So what I am assuming to do is first break it apart as in

Ca + H₂  CaH₂

and afterward i don't know where to go from there, what should I due next?

[Big-Daddy]

Calculate the lattice energy for CaH₂, in kJ/mole, using the following information:
EA for H = -73 kJ/mol; BE for hydrogen = 432 kJ/mol
IE₁ for Ca = 591 kJ/mol; IE₂ for Ca = 1146 kJ/mol
ΔH_f for CaH₂ = -186 kJ/mol


So what I am assuming to do is first break it apart as in

Ca + H₂  CaH₂

and afterward i don't know where to go from there, what should I due next?

You're missing a needed piece of information - the enthalpy change of atomisation of Ca (s). You need to convert Ca (s) to gas before you can use its ionization energies.

Give me that and I could get you the lattice formation enthalpy.

[Superboy92]

well that's all what the info that was given, i was told that i am suppose to change Ca(s) to (g)
So I'm assuming that I have to do something like this:

Ca(s)  Ca(g)

and

H₂ (g)  2H(g)

and then I'm stuck again

[Big-Daddy]

well that's all what the info that was given, i was told that i am suppose to change Ca(s) to (g)
So I'm assuming that I have to do something like this:

Ca(s)  Ca(g)

and

H₂ (g)  2H(g)

and then I'm stuck again

Without the enthalpy change of the first reaction (which is the correct reaction) I can't see how you would find the lattice enthalpy, but perhaps someone else knows.

You do have the enthalpy change of the second reaction already - it is simply the bond dissociation enthalpy of a H-H bond, which according to your data is +432 kJmol^-1.

[SinkingTako]

Without the enthalpy change of the first reaction (which is the correct reaction) I can't see how you would find the lattice enthalpy, but perhaps someone else knows.

currently, I'm also facing the same problem. I was given quite a few problems to find the enthaphy of reaction, where energy of atomisation was not provided. so any help in this area (predicting △H without using △H_atm ) will be greatly appreciated too! Thanks!

[Big-Daddy]

Without the enthalpy change of the first reaction (which is the correct reaction) I can't see how you would find the lattice enthalpy, but perhaps someone else knows.

currently, I'm also facing the same problem. I was given quite a few problems to find the enthaphy of reaction, where energy of atomisation was not provided. so any help in this area (predicting △H without using △H_atm ) will be greatly appreciated too! Thanks!

I'm not convinced it's possible, so please ask your Professor and do get back to me, I'd be interested to know.

[antimatter101]

Does BE stand for Bejan Number? If I know that, possibly I can solve it.

[Big-Daddy]

Does BE stand for Bejan Number? If I know that, possibly I can solve it.

Haha somewhat optimistic! I'm confident BE stands for bond enthalpy, or more accurately bond dissociation enthalpy. In this case it is the energy taken in to break the mean H-H bond.

Is it then possible? I'm convinced you cannot work this out with the enthalpy change Ca (s) -> Ca (g) (enthalpy change of atomization of Ca (s)).

[antimatter101]

I just checked, and the enthalpy of atomisation for Ca is 178kj/mol. So...

167 - 591 - 1146 + 432 - 73 - 73 - 186 = -1470kj/mol

[Big-Daddy]

I just checked, and the enthalpy of atomisation for Ca is 178kj/mol. So...

167 - 591 - 1146 + 432 - 73 - 73 - 186 = -1470kj/mol

You've got a +167 coming in from seemingly nowhere, you're missing the -178 (enthalpy change of atomisation may be positive but you must subtract it in the calculation for lattice formation enthalpy), you've got +432 in there when you should have -432 (again, enthalpy change of bond dissociation may be positive but you have to subtract it in this calculation). I got -2,387 kJmol-1 using +178 for the enthalpy change of atomisation of Ca (s), but I'm still waiting to hear that there is a way to solve it without this enthalpy change. 

[antimatter101]

I was hurrying in my reply coz i had other stuff to do, that is why i made so many mistakes.

Anyway I think there may be a solution to "how to calculate lattice energy without enthalpy of atomization" on this website:

http://www.chemguide.co.uk/physical/energetics/lattice.html

or

http://www.chemguide.co.uk/physical/energetics/sums.html

Have a nice day!

[Big-Daddy]

I was hurrying in my reply coz i had other stuff to do, that is why i made so many mistakes.

Anyway I think there may be a solution to "how to calculate lattice energy without enthalpy of atomization" on this website:

http://www.chemguide.co.uk/physical/energetics/lattice.html

or

http://www.chemguide.co.uk/physical/energetics/sums.html

Have a nice day!

Nope, there isn't! That is the website I first learnt energetics from. It's a website for UK A-Level pupils and is not up to this problem. It would describe the general method we need if the enthalpy of atomisation was there, but as it stands we can do nothing. Look at your Hess law equations: you can't go anywhere without Ca (s) -> Ca (g)!

If there is a solution to this it is above and beyond Hess' law.