Question

What is the change in entropy when 0.135 mol of potassium freezes at 62.6

Answer 1

Assuming that the freezing potassium is in equilibrium with melting potassium, then DG (Gibbs free energy) is zero. DG is zero for a system at equilibrium.

When freezing takes place the DH is negative since heat is being released, and since the substance is going to a solid the entropy must be decreasing.

DGo = DHo -TDSo
0 = DHo - TDSo
Then
DHo = TDSo
or
DSo = DHo / T
DSo = (-2.39 kJ / mol) / 340.8 K = -0.00701 kJ/mol K
or
-7.01 J/mol K

Then for 0.135 mol ....
DS = (-7.01 J/mol K) x 0.135 mol = -0.94635 J/K Ans.