As O₂(l)is cooled at 1 atm,it freezes at 54.5 K to form Solid I.At a lower temperature,Solid I rearranges to Solid II,which has a different crystal structure.Thermal measurements show that $\Delta$H for the I $\to$II phase transition is -743.07 J/mol,and $\Delta$S for the same transition is -17.0 J/K mol.At what temperature are Solids I and II in equilibrium?

Question

Quizplus · Accepted Answer

At the phase transition, the change in Gibbs free energy (∆G) is equal to zero (since the two phases are in equilibrium). Using the relationship ∆G = ∆H - T∆S, we can solve for the temperature at which ∆G = 0. 
0 = ∆H - T∆S 
T = ∆H/∆S 
T = (-743.07 J/mol) / (-17.0 J/K mol) = 43.7 K 
Therefore, solids I and II are in equilibrium at 43.7 K.

As O2(l)is Cooled at 1 Atm,it Freezes at 54 Δ\DeltaΔ H for the I

As O₂(l)is Cooled at 1 Atm,it Freezes at 54 $\Delta$ H for the I