Use the data given below to construct a born-haber cycle to determine the electron affinity of br. △ h°(kj) k(s) → k(g) 89 k(g) → k+(g) + e- 419 br2(l) → 2 br(g) 193 k(s) + 12 br2(g) → kbr(s) -394 kbr(s) → k+(g) + br-(g) 674 i know the answer is -325 kj, i want to know how to get it.

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