HELPPPP! Entropy challenge question!?

Question

The Delta(G) for the freezing of H2O(l) at -10 degrees celsius is -210 j/mol and the heat of fusion of ice at this temperature is 5610 J/mol.

Answer 1

As water freezes at -10C heat is released, so DH is negative.

DG = DH - TDS

DG - DH = -TDS

DS = (DG - DH) / -T

DS = (-210 J/mol + 5610 J/mol) / -263K



DS = -20.5 J / molK

========= Follow up ==========

DH = TDS is only true when DG is equal to zero and that is only true when a system is at equilibrium.  An equilibrium between ice and water exists at the freezing point (0C), not at -10C.

Answer 2

delS = delH/T  -10 C + 273.15 = 263.15K

         = - 5610/263.15  J/(mol-K) = - 21.31J/(mol-K)

Answer 3

delS = delH/T -10 C + 273.15 = 263.15K

= - 5610/263.15 J/(mol-K) = - 21.31J/(mol-K)