Question

Τhe enthalpy of vaporization of ammonia is 23.35 kJ/mol at its boiling point (–33.4°C). Calculate the value of ∆S when 1.00 mole of ammonia is vaporized at –33.4°C and 1.00 atm

Answer 1

Answer:

ΔS = 0.0975 KJ/K

Explanation:

Trouton's rule relates the enthalpy and entropy of vaporization at the normal boiling point.

• ΔSvap = ΔHvap / Tbp

∴ Tbp = - 33.4°C ≅ 239.5 K

∴ ΔHvap = 23.35 KJ/mol

⇒ ΔS = (23.35 KJ/mol) / (239.5 K) = 0.0975 KJ/mol.K

for 1 mol of NH3:

⇒ ΔS = 0.0975 KJ/K