Need help with the Born-Haber Cycle?

Question

Calculate the missing information for the ionic solid formation from the elements starting in their elemental states.

 

Ba(s) + 1/2 O2(g) ---> BaO(s)

 

Determine the second ionization energy of Barium. The first ionization energy is 503 kJ/mol; sublimation energy is 180 kJ/mol; the lattice energy is 3,125 kJ/mol; the net reaction energy is -548 kJ/mol; the bond dissociation of oxygen is 498 kJ/mol; the first electron affinity of oxygen is -141 kJ/mol and the second is 744 kJ/mol. Each individual equation needs to be balanced.

 

I know how to do the cycle but I would like to know where the numbers go for this particular problem. This problem is just worded so awful that I just cannot understand it. I would really appreciate any help I can get. Thank you!

Answer 1

Hello Emily,

Below you can see how this reaction breaks down to its individual constituents:

Ba(s)→Ba(g) ΔHsub​=180 kJ/mol

Ba(g)→Ba+(g)+e− ΔHIE1​​=503 kJ/mol

Ba+(g)→Ba2+(g)+e− ΔHIE2​​=?

0.5 ​O2​(g)→O(g) 0.5*498​=249 kJ/mol

O(g)+e−→O−(g) ΔHEA1​​=−141 kJ/mol

O−(g)+e−→O2−(g) ΔHEA2​​=744 kJ/mol

Ba2+(g)+O2−(g)→BaO(s) ΔHlattice​=−3125 kJ/mol

ΔHnet​=−548 kJ/mol

If you plug in the values, you can find the solution for IE2.

ΔHnet​=ΔHsub​+ΔHIE1​​+ΔHIE2​​+ΔHdiss​+ΔHEA1​​+ΔHEA2​​+ΔHlattice​

−548=180+503+ΔHIE2​​+249+(−141)+744+(−3125)

ΔHIE2​​=842 kJ/mol

If you write out the entire equation, the second ionization energy should equal to 842 kJ/mol.