This is the value for the electron affinity = -339.8 kJ
Review the problem because it is possibly wrong and there are also incomplete or erroneous data
Explanation:
First of all, you have to think the chemical reaction, based on the elements in their ground state.
K(g) + 1/2 Br₂ (l) → KBr
How do we find bromine or potassium in nature? Br₂ as gas, K as liquid.
For this reaction, we use △Hf (kJ) = -394 (formation enthalpy)
The reaction is then defined from the elements in the gaseous state, to form the crystals of the salt, so Br and K have to change state. At the end, the equation will be:
K⁺(g) + Br⁻(g) → KBr This process used the energy called, lattice energy.
LE = -674 kJ.
So we have to go, from K(s) to K⁺(g), and from Br₂(l) to Br⁻(g).
First of all, we have to convert K(s) → K(g) with △Hsublimation: 89kJ
And then tear out an electron to form the cation, with the ionization energy K(g) → K⁺(g) + 1e⁻ △H: 419 kJ
In first place, we have to convert Br₂(l) to Br₂(g) with a vaporization process. For this: Br₂(l) → Br₂(g) △H: 30.7 kJ (THIS VALUE IS MISSING AND IT IS WRONG IN WHAT YOU WROTE)
Notice we have, a half of 1 mol of bromine, so we have to convert a half of 1 mol, so we need a half of energy. The enthalpy vaporization is for 1 mol of Br₂, but we only have a half.
Aftewards, we have to separate the 1/2Br₂(g). As this is a dyatomic molechule, we need only 1 Br.
DEFINETALY THERE IS MISTAKE ON WHAT YOU WROTE BECAUSE THIS VALUE IS INCORRECT IN THE STATEMENT.
You use the enthalpy for dissociation to have this Br-Br. You must break the bond. △H = 193/2 kJ
And as you have 1/2 mol, you need 1/2 of energy
Now we have to apply, the electron affinity, to get the bromide anion.
1/2Br₂(g) + 1e- → Br⁻ (g) △H: ?
This is the unknown value.
How do you make the Born Haber cycle? The Sum all the △H + LE = △Hf
LE + △Hs + △Hie + △Hv + △Hdis + EA = -394 kJ
EA = -394kJ - LE - △Hs - △Hie - △Hv - △Hdis
EA = -394kJ + 674 kJ - 89kJ - 419 kJ - 30.7/2 kJ - 193/2 kJ
EA = -339.8 kJ
