Lecture_05

# This yields dh 1 1 ln x a fus m t t 5115 r f we

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: d and solid phases, we require µlA =µsA, or l ,l + RT ln x A = l s, from which we get A A l s − l ,l −D fus G A,m , at constant T and P. ln x A = A RT A = RT Differentiating both sides with respect to T, we get éæ öù D fus H m d ln x A −1 ê ¹ è D fus G A,m /T ø ú = Rê ú = RT 2 , at the equilibrium temperature. dT ¹T ê ú ë ûP Now we rearrange and integrate both sides, the LHS from a mole fraction of xA = 1 (pure solvent) to the solvent mole fraction in solution, xA < 1, and correspondingly, the RHS from a liquid-solid equilibrium temperature of T*f (freezing point of the pure solvent) to T, the freezing point of the solution. This yields DH 1 1 ln x A = fus m æ T  − T ö . ÷ (5.115) R çf è ø We may re-write this as D fus H m æ T − T  ö ln(1 − x B ) = R ç T  T f ÷ . èf ø (5.116) We now make two simplifications: for small xB, ln(1–xB) ≈ –xB (see Eq. 5.117) and T*f Tf ≈ T*f2. This gives D H æ DT ö x B = fus m ç 2f ÷ , R è Tf ø which can be further simplified...
View Full Document

## This document was uploaded on 02/28/2014 for the course CHEM 311 at LA Tech.

Ask a homework question - tutors are online