Question

(P48.5 Calculate the difference in slope of the chemical potential against tem- perature on either side of (a) the normal freezing point of water and (b) the normal boiling point of water. The molar entropy change accompanying fu- sion is 22.0JK 'mol"' and that accompanying evaporation is 109.9JK 'mol", (c) By how much does the chemical potential of water supercooled to -5.0°C exceed that of ice at that temperature?

Answer 1

At the transition from phase ato phase , we have In which represent represents the difference in the slopes of the chemical potentials of the phase at either side of the phase transition. (a) For, liquid solid phase change the above equation becomes (), 4) --- +5.-45-7 From table 23: A H=6.011 Jmol =-22.00 Jmol 'K Far, liquid + vapor phaz change (5) From table 23: 4_H=40.61kJmol* -40.6kJmol 373.15K =108.8.JKmol

c) We know that Aue=( BATSAAT A4-44 = (442-Mare)-(4.42-Hore) = thre-thuc-hscthuc At0°C, liquid and solid are in one phase de ll e So A4-A4 = 440-44 (S..,-..) AT -ASAT =-(-22.0JK-'mol-")x SK =+110.0Jmol' Since Afie > Alici thoro is a thermodynamic to ridency to from