Question

The normal boiling point of c2cl3f3 is 47.6°c and its molar enthalpy of vaporization is 27.49 kj/mol. what is the change in entropy in the system in j/k when 24.1 grams of c2cl3f3 vaporizes to a gas at the normal boiling point?

Answer 1

According to this formula when:

ΔG = ΔH - TΔS = 0

∴ ΔS = ΔH/T

∴ ΔS = n*ΔHVap / Tvap

- when n is the number of moles = mass/molar mass 

when the mass = 24.1 g 

and the molar mass = 187.3764 g/mol

by substitution: 

∴ n = 24.1 / 187.3764g/mol

      = 0.129 moles

and ΔHvap is the molar enthalpy of vaporization is 27.49 kJ/mol

and Tvap is the temperature in Kelvin = 47.6 + 273 = 320.6 K

So by substitution, we will get the ΔS the change in entropy:

∴ΔS = 0.129 mol * 27490 J/mol / 320.6 K

      = 11 J/K