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 $\rightarrow$II phase transition is -743.1 J/mol and that $\Delta$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

The equilibrium temperature can be found using the formula ΔG = ΔH - TΔS = 0. Solving for T gives T = ΔH/ΔS = (-743.1 J/mol) / (-17.0 J/K mol) = 43.7 K.

As O2(l) Is Cooled at 1 Atm, It Freezes at 54.5

As O₂(l) Is Cooled at 1 Atm, It Freezes at 54.5