Question

The normal boiling point of methanol is 64.7 ∘ C and the molar enthalpy of vaporization is 71.8kJ/mol . The value of ΔS when 1.35 mol of CH 3 OH(l) vaporizes at 64.7 ∘ C is ________ J/K .

Answer 1

Answer:

286 J/K

Explanation:

The molar Gibbs free energy for the vaporization (ΔGvap) is:

ΔGvap = ΔHvap - T.ΔSvap

where,

ΔHvap: molar enthalpy of vaporization

T: absolute temperature

ΔSvap: molar entropy of the vaporization

When T = Tb = 64.7 °C = 337.9 K, the reaction is at equilibrium and ΔGvap = 0.

ΔHvap - Tb . ΔSvap = 0

ΔSvap = ΔHvap/Tb = (71.8 × 10³ J/K.mol)/ 337.9 K = 212 J/K.mol

When 1.35 mol of methanol vaporizes, the change in the entropy is:

$1.35mol.\frac{212J}{K.mol} =286 J/K$