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

ΔS = 0.0975 KJ/K

Explanation:

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

∴ 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

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

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