Question

The normal boiling point of a substance is defined to be the temperature at which the liquid phase of the substance is in equilibrium with the gas phase at 1 atm pressure. The normal boiling point of methanol is 80oC and ∆H vap = 38 kJ/mol. What is the ∆Scrap value

Answer 1

ΔSv = 0.1075 KJ/mol.K

Step-by-step explanation:

Binary solution:

∴ a: solvent

∴ b: solute

in equilibrium:

• μ*(g) = μ(l) = μ* +RTLnXa....chemical potential (μ)

⇒ Ln (1 - Xb) = ΔG/RT

∴ ΔG = ΔHv - TΔSv

⇒ Ln(1 -Xb) = ΔHv/RT - ΔSv/R

∴ Xb → 0:

⇒ Ln(1) = ΔHv/RT - ΔSv/R

∴ T = T*b....normal boiling point

⇒ 0 = ΔHv/RT*b - ΔSv/R

⇒ ΔSv = (R)(ΔHv/RT*b)

⇒ ΔSv = ΔHv/T*b

∴ T*b = 80°C ≅ 353 K

⇒ ΔSv = (38 KJ/mol)/(353 K)

⇒ ΔSv = 0.1075 KJ/mol.K