Question

If lithium has a first ionisation energy of `520` `kJ` `mol^(-1)`, estimate the energy required to form 1 `Li^+` ion from 1 Li atom?

Answer 1

Around `8.64*10^(-19)J=5.40eV`

Explanation:

The ionization energy tells us that for 1 mole of gaseous lithium, `520kJ` of energy is needed to ionise each atom.

We also know that 1 mole contains `6.02*10^23` atoms (to `3` s.f.)

So, 1 atom = `1/(6.02*10^23)=6.02*10^(-23) mol`

`520kJ=520000J=5.2*10^5J`

`(5.2*10^5)(6.02*10^(-23))=8.64*10^(-19)J`

`(8.64*10^(-19))/(1.60*10^(-19))=5.40eV`

Answer 2

We are given the first ionization energy of the lithium atom....

Explanation:

That is we are given data for the reaction:

`Li(g) +Delta rarr Li^+(g)+e^(-)`...

Where `Delta=520*kJ*mol^-1`, and when we write `kJ*mol^-1` we mean PER MOLE of reaction as written.....

And given the equation we need `520*kJ` to produce a mole of gaseous lithium cations....

To produce a SINGLE LITHIUM CATION, we divide the molar quantity by the Avocado number....the number of avocadoes in a mole, `N_A=6.022xx10^23*mol^-1`.

`(520*kJ*mol^-1)/(6.022xx10^23*mol^-1)=8.64xx10^-22*kJ=8.64xx10^-19*J`. We could convert this value to `"ergs"` or `cm^-1` or `eV` if we are real masochists.