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

Answer:

83.72 K

Explanation:

$\alpha$ = Polarizability of argon = $1.66\times 10^{-30}\ m^3$

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

$P=-\frac{3}{4}\frac{\alpha^2I}{r^6}\\\Rightarrow P=-\frac{3}{4}\frac{(1.66\times 10^{-30})^2\times 1521\times 10^3}{(3.8\times 10^{-10})^6}\\\Rightarrow P=-1044.01\ J/mol$

Kinetic energy is given by

$K=\frac{3}{2}RT$

The potential and kinetic energy will balance each other

$P=\frac{3}{2}RT\\\Rightarrow 1.04401\times 10^{-33}=\frac{3}{2}RT\\\Rightarrow T=\frac{1044.01\times 2}{3\times 8.314}\\\Rightarrow T=83.72\ K$

The boiling point of argon is 83.72 K