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?

2 Answers
Nov 15, 2017

Answer:

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#

Nov 15, 2017

Answer:

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.