Question

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 1

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.