Lecture_05

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

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

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