Question

The internuclear distance between two closest Ar atoms in solid argon is about 3.8 A. The polarizability of argon is 1.66e-30 m3, and the first ionization is 1521kJ/mol. Estimate the boiling point of argon [Hint: Calculate the potential energy due to dispersion interaction for solid argon, and equate this quantity to the average kinetic energy of 1mole of argon gas, which is (3/2)RT].

I'm confused about where to use the first inonization energy?

Answer 1

83.72 K

Step-by-step explanation:

`\alpha` = Polarizability of argon =
`1.66* 10^(-30)\ m^3`

I = First ionization = 1521 kJ/mol

r = Distance between atoms = 3.8 A

R = Gas constant = 8.314 J/mol K

T = Boiling point

Potential energy due to dispersion of gas is given by

`P=-(3)/(4)(\alpha^2I)/(r^6)\\\Rightarrow P=-(3)/(4)((1.66* 10^(-30))^2* 1521* 10^3)/((3.8* 10^(-10))^6)\\\Rightarrow P=-1044.01\ J/mol`

Kinetic energy is given by

`K=(3)/(2)RT`

The potential and kinetic energy will balance each other

`P=(3)/(2)RT\\\Rightarrow 1.04401* 10^(-33)=(3)/(2)RT\\\Rightarrow T=(1044.01* 2)/(3* 8.314)\\\Rightarrow T=83.72\ K`

The boiling point of argon is 83.72 K