If lithium has a first ionisation energy of 520520 kJkJ mol^(-1)mol1, estimate the energy required to form 1 Li^+Li+ ion from 1 Li atom?

2 Answers
Nov 15, 2017

Around 8.64*10^(-19)J=5.40eV8.641019J=5.40eV

Explanation:

The ionization energy tells us that for 1 mole of gaseous lithium, 520kJ520kJ of energy is needed to ionise each atom.

We also know that 1 mole contains 6.02*10^236.021023 atoms (to 33 s.f.)

So, 1 atom = 1/(6.02*10^23)=6.02*10^(-23) mol16.021023=6.021023mol

520kJ=520000J=5.2*10^5J520kJ=520000J=5.2105J

(5.2*10^5)(6.02*10^(-23))=8.64*10^(-19)J(5.2105)(6.021023)=8.641019J

(8.64*10^(-19))/(1.60*10^(-19))=5.40eV8.6410191.601019=5.40eV

Nov 15, 2017

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.