The energy required to dissociate KF into neutral atoms is 498 kJ/mol. Given that the first ionization energy for K is 418 kJ/mol, calculate the electron affinity (in kJ/mol) for F. Show your work for all calculations.

Answer: Electron affinity of F equals

275.8kJ/mol

Explanation: Electron affinity is the energy change when an atom gains an electron.

Let's first calculate the energy required - E (r) to dissociate KF into ions not neutral atom which is given.

E (r) = {z1*z2*e²}/{4π*permitivity of space*r}

z1 is - 1 for flourine

z2 is + 1 for potassium

e is magnitude of charge 1.602*EXP{-9}C

r is ionic bond length of KF (is a constant for KF 0.217nm)

permitivity of free space 8.854*EXP{-12}.

Now let's solve

E (r) = { (-1) * (1) * (1.602*EXP{-9}) ²} /

{4*3.142*8.854*EXP (-12) * 0.217*EXP (-9)

E (r) = - 1.063*EXP{-18}J

But the energy is released out that is exothermic so we find - E (r)

Which is + 1.603*EXP{-18}J

Let's now convert this into kJ/mol

By multiplying by Avogadro constant 6.022*EXP (23) for the mole and diving by 1000 for the kilo

So we have,

1.603*EXP (-18) * 6.022*EXP (23) / 1000

-E (r) = 640.2kJ/mol.

Now let's obtain our electron affinity for F

We use this equation

Energy of dissociation (nuetral atom) = electron affinity of F + (-E (r)) + ionization energy of K.

498kJ/mol

=e affinity of F + 640.2kJ/mol

+ (-418kJ/mol)

(Notice the negative sign in ionization energy for K. since it ionize by losing an electron)

Making electron affinity of F subject of formula we have

Electron affinity (F) = 498+418-640.2

=275.8kJ/mol.