Question

A different derivation of the freezing point depression begins with the expression

ln x_A = −∆fusG/RT
and uses the Gibbs-Helmholtz equation. Make the substitution and integrate over appropriate limits to derive the freezing point depression equation. Be sure to state any assumptions you make.

Answer 1

x_A = mole fraction of the solvent

x_B = mole fraction of the solute

ln x_A = - Δ_fusG/RT

or, ln (1- x_B) = - Δ_fusG/RT

or, d ln (1- x_B) = - (1/R) d (Δ_fusG/T)

or, d ln (1- x_B) = (1/R) (ΔH_fus/T² )dT

or, ₀∫^xB d ln (1- x_B) = (1/R) _Tf∫^T (ΔH_fus/T² ) dT

or, [ln (1- x_B)]₀^XB = (ΔH_fus /R) ( - 1/T)^T_Tf

or, ln (1-x_B) = - (ΔH_fus /R)(1/T - 1/T_f)

or, -x_B = - (ΔH_fus /R)(1/T - 1/T_f)                    (since for a dilute solution x_B << 1, so ln (1-x_B) can be written as -                                                                                    x_B)

or, x_B = (ΔH_fus /RTT_f)(T_f - T)

= ΔH_fus ΔT_f /RTT_f

= ΔH_fus ΔT_f /RT_f²                     (as T is very close to T_f)