Question

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.

Answer 1

Answer:

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

Answer 2

Answer : The electron affinity of Br is, -324.5 kJ

Explanation :  

The formation of potassium bromide is,

$K^{1+}(g)+\frac{1}{2}Br_2(g)\overset{\Delta H_L}\rightarrow KBr(s)$

$\Delta H_f^o$ = enthalpy of formation of potassium bromide

The steps involved in the born-Haber cycle for the formation of $KBr$:

(1) Conversion of solid lithium into gaseous potassium atoms.

$K(s)\overset{\Delta H_s}\rightarrow K(g)$

$\Delta H_s$ = sublimation energy of potassium

(2) Conversion of gaseous potassium atoms into gaseous potassium ions.

$K(g)\overset{\Delta H_I}\rightarrow K^{+1}(g)$

$\Delta H_I$ = ionization energy of potassium

(3) Conversion of molecular gaseous bromine into gaseous bromine atoms.

$Br_2(g)\overset{\Delta H_D}\rightarrow 2Br(g)$

$\frac{1}{2}Br_2(g)\overset{\Delta H_D}\rightarrow Br(g)$

$\Delta H_D$ = dissociation energy of bromine

(4) Conversion of gaseous bromine atoms into gaseous bromine ions.

$Br(g)+e^-\overset{\Delta H_E}\rightarrow Br^-(g)$

$\Delta H_E$ = electron affinity energy of bromine

(5) Conversion of gaseous cations and gaseous anion into solid potassium bromide.

$K^{1+}(g)+Br^-(g)\overset{\Delta H_L}\rightarrow KBr(s)$

$\Delta H_L$ = lattice energy of potassium bromide

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I+\frac{1}{2}\Delta H_D+\Delta H_E+\Delta H_L$

Now put all the given values in this equation, we get:

$-394=89+419+\frac{1}{2}\times 193+\Delta H_E+(-674)$

$\Delta H_E=-324.5kJ$

Therefore, the electron affinity of Br is, -324.5 kJ